Ionic Equilibrium

Given $pK_a = 4$, so

$K_a = 10^{-pK_a} = 10^{-4}$

For a weak acid solution of concentration $C = 10\,\text{mM} = 0.01\,\text{M}$:

Let the dissociation be $x$:

$\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^-$

At equilibrium:

• $[\mathrm{H}^+] = x$

• $[\mathrm{A}^-] = x$

• $[\mathrm{HA}] = C - x \approx C$ (given $x$ is negligible compared to $C$)

So,

$K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} \approx \frac{x^2}{C}$

Therefore,

$x = \sqrt{K_a C} = \sqrt{(10^{-4})(10^{-2})} = \sqrt{10^{-6}} = 10^{-3}$

So,

$[\mathrm{H}^+] = 10^{-3}\,\text{M} \Rightarrow pH = -\log(10^{-3}) = 3$

Nearest integer = $\boxed{3}$

$ \mathrm{XS}(\mathrm{~s}) \rightleftharpoons \mathrm{X}^{+2} \text { (aq.) }+\mathrm{S}^{2-} \text { (aq.) } $

For precipitation of $\mathrm{XS}(\mathrm{s})$, we use the solubility product condition:

$ \begin{aligned} & {\left[\mathrm{X}^{+2}\right]\left[\mathrm{S}^{2-}\right] \geq \mathrm{K}_{\mathrm{sp}}(\mathrm{XS})} \\ & \text{Given } [\mathrm{X}^{2+}] = 0.01 \text{ M, so } \left[\mathrm{S}^{2-}\right] \geq \frac{1 \times 10^{-22}}{0.01}=10^{-20} \\ & \mathrm{YS}(\mathrm{~s}) \rightleftharpoons \mathrm{Y}^{+2}(\mathrm{aq})+\mathrm{S}^{2-}(\mathrm{aq}) \end{aligned} $

For precipitation of $\mathrm{YS}(\mathrm{s})$, again apply $K_{sp}$ condition:

$ \begin{aligned} & {\left[\mathrm{Y}^{+2}\right]\left[\mathrm{S}^{2-}\right] \geq \mathrm{K}_{\mathrm{sp}}(\mathrm{YS})} \\ & \text{Given } [\mathrm{Y}^{2+}] = 0.01 = 10^{-2} \text{ M, so } \left[\mathrm{S}^{2-}\right] \geq \frac{4 \times 10^{-16}}{10^{-2}}=4 \times 10^{-14} \end{aligned} $

Now we find how much $\mathrm{S}^{2-}$ comes from $\mathrm{H}_2\mathrm{S}$. In water:

Now, $\mathrm{H}_2 \mathrm{~S}(\mathrm{aq}) \rightleftharpoons 2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{S}^{2-}(\mathrm{aq})$

Using the given relation $\mathrm{K}_{a1}\times \mathrm{K}_{a2}=1 \times 10^{-21}$:

$\begin{aligned} & \frac{\left[\mathrm{S}^{2-}\right]\left[\mathrm{H}^{+}\right]^2}{\mathrm{H}_2 \mathrm{~S}}=\mathrm{K}_{\mathrm{a}_1} \times \mathrm{K}_{\mathrm{a}_2}=1 \times 10^{-21} \\ & \Rightarrow {\left[\mathrm{~S}^{2-}\right]=\frac{1 \times 10^{-21} \times\left[\mathrm{H}_2 \mathrm{~S}\right]}{\left[\mathrm{H}^{+}\right]^2}} \end{aligned}$

For $\mathrm{YS}$ to start precipitating, we need $\left[\mathrm{S}^{2-}\right] \geq 4 \times 10^{-14}$. So:

$\begin{aligned} & \frac{1 \times 10^{-21} \times\left[\mathrm{H}_2 \mathrm{~S}\right]}{\left[\mathrm{H}^{+}\right]^2} \geq 4 \times 10^{-14} \\ & \text{Given } [\mathrm{H}_2\mathrm{S}] = 0.1 = 10^{-1} \text{ M} \\ & \Rightarrow \frac{1 \times 10^{-21} \times 10^{-1}}{\left[\mathrm{H}^{+}\right]^2} \geq 4 \times 10^{-14} \\ & \Rightarrow \left[\mathrm{H}^{+}\right]^2 \leq \frac{10^{-22}}{4 \times 10^{-14}}=\frac{1}{4}\times 10^{-8} \\ & { \left[\mathrm{H}^{+}\right]^2 \leq \frac{1}{4} \times 10^{-7} \times 10^{-1}} \\ & \Rightarrow { \left[\mathrm{H}^{+}\right] \leq \frac{1}{2} \times 10^{-4}} \\ & \Rightarrow \mathrm{pH} \geq 4.3 \end{aligned}$

For the diprotic acid $ \mathrm{H_2X} $ (total concentration $C=0.1\,\mathrm{M}$):

$ K_{a1}=2.5\times 10^{-8},\qquad K_{a2}=1.0\times 10^{-13} $

1) Find $[\mathrm{H^+}]$ (second dissociation is negligible for $[\mathrm{H^+}]$):

$ [\mathrm{H^+}] \approx \sqrt{K_{a1}C} = \sqrt{(2.5\times 10^{-8})(0.1)} = \sqrt{2.5\times 10^{-9}} = 5.0\times 10^{-5}\,\mathrm{M} $

2) Use diprotic distribution for $[\mathrm{X^{2-}}]$:

$ [\mathrm{X^{2-}}]= C\cdot \frac{K_{a1}K_{a2}/[\mathrm{H^+}]^2}{1+K_{a1}/[\mathrm{H^+}]+K_{a1}K_{a2}/[\mathrm{H^+}]^2} $

Compute key terms:

$ \frac{K_{a1}K_{a2}}{[\mathrm{H^+}]^2} =\frac{(2.5\times 10^{-8})(10^{-13})}{(5\times 10^{-5})^2} =\frac{2.5\times 10^{-21}}{2.5\times 10^{-9}} =10^{-12} $

$ \frac{K_{a1}}{[\mathrm{H^+}]}=\frac{2.5\times 10^{-8}}{5\times 10^{-5}}=5\times 10^{-4} $

So denominator $\approx 1+5\times 10^{-4}\approx 1.0005$, hence

$ [\mathrm{X^{2-}}] \approx 0.1 \times \frac{10^{-12}}{1.0005} \approx 1.0\times 10^{-13}\,\mathrm{M} $

Expressed as $\_\_\_\_ \times 10^{-15}\,\mathrm{M}$:

$ 1.0\times 10^{-13}\,\mathrm{M} = 100\times 10^{-15}\,\mathrm{M} $

Nearest integer = $\boxed{100}$.

$\begin{aligned} &\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\ & \mathbf{p O H}=4.745 \end{aligned}\\ &\text { on adding } 0.05 \text { mole } \mathrm{HCl} \end{aligned}$

$\begin{array}{llll} \mathrm{NH}_3+ & \mathrm{H}^{\oplus} & \rightarrow & \mathrm{NH}_4^{\oplus} \\ 0.1 & 0.05 & & 0.1 \\ 0.05 & 0 & & 0.15 \end{array}$

$\begin{aligned} & \mathrm{pOH}^{\prime}=4.745+\log 3 \\ & \mathrm{pOH}^{\prime}-\mathrm{pOH}=0.477 \\ & 14-\mathrm{pH}^{\prime}-14+\mathrm{pH}=0.477 \\ & \Delta \mathrm{pH}=0.477 \\ & =47.7 \times 10^{-2} \approx 48 \times 10^{-2} \end{aligned}$

To determine the percentage dissociation of the salt $\text{MX}_3$, consider the following dissociation reaction:

$ \text{MX}_3 \rightarrow \text{M}^{+3} + 3\text{X}^{-} $

The van't Hoff factor, $\text{i}$, is given as 2. The formula relating $\text{i}$ to the degree of dissociation, $\alpha$, is:

$ \text{i} = 1 + (\text{n} - 1) \alpha $

Here, $\text{n}$ represents the total number of ions produced per formula unit of the salt. For $\text{MX}_3$, dissociation yields:

1 $\text{M}^{+3}$ ion

3 $\text{X}^{-}$ ions

Thus, $\text{n} = 4$. Substituting into the formula, we have:

$ 2 = 1 + (4 - 1) \alpha $

Simplifying, we solve for $\alpha$:

$ 2 = 1 + 3\alpha $

$ 3\alpha = 1 $

$ \alpha = \frac{1}{3} \approx 0.3333 $

As a percentage, this degree of dissociation is:

$ \alpha \times 100\% \approx 33.33\% $

Rounding to the nearest integer gives:

$ 33\% $

Therefore, the percentage dissociation of the salt $\text{MX}_3$ is approximately 33%.

$\begin{aligned} & \lambda_{\mathrm{m}}^{\circ} \text { of } \mathrm{NH}_4 \mathrm{Cl}=185 \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{+}}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Cl}^{-}}=185 \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{+}}=185-70=115 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}=\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4^{+}}+\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{OH}^{-}} \\ & =115+170 \\ & \left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}=285 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \end{aligned}$

$\begin{aligned} \text { degree of dissociation } & =\frac{\left(\lambda_{\mathrm{m}}\right)_{\mathrm{NH}_4 \mathrm{OH}}}{\left(\lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{NH}_4 \mathrm{OH}}} \\ & =\frac{85.5}{285} \\ & =0.3 \\ & =3 \times 10^{-1} \end{aligned}$

To determine the mass of Mg(OH)₂ that needs to be dissolved to achieve a pH of 10.0, follow these steps:

Given:

pH = 10

Therefore, pOH = 14 - pH = 4

[OH⁻] = 10⁻⁴ M

First, calculate the number of moles of OH⁻ ions:

Number of moles of OH⁻ = 10⁻⁴

Since Mg(OH)₂ dissociates completely in water, from one mole of Mg(OH)₂, you get two moles of OH⁻:

Number of moles of Mg(OH)₂ = (10⁻⁴) / 2 = 5 × 10⁻⁵ moles

Next, calculate the mass of Mg(OH)₂:

Molar mass of Mg(OH)₂ = 58 g/mol

Convert the number of moles to mass:

Mass of Mg(OH)₂ = (5 × 10⁻⁵ moles) × (58 g/mol) = 2.9 × 10⁻³ g

Convert to milligrams: 2.9 × 10⁻³ g = 2.9 mg

Therefore, the calculated mass of Mg(OH)₂ required is approximately 2.9 mg.

$\mathrm{HX}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}$

$\begin{aligned} &0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad \text { Not justified }\\ &\begin{aligned} & \Rightarrow 0.01 \alpha=10^{-5} \Rightarrow \alpha=10^{-3} \\ & \mathrm{~K}_{\mathrm{a}}=0.01 \alpha^2=10^{-8} \end{aligned} \end{aligned}$

$\begin{aligned} &\text { On dilution let final concentration of } \mathrm{HX}=\mathrm{c} \mathrm{M}\\ &\mathrm{Hx}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \end{aligned}$

$\mathrm{C}(1-\alpha) \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha$

$\begin{aligned} \Rightarrow & C \alpha=10^{-6} \quad\text{..... (1)}\\ & \frac{C \alpha^2}{1-\alpha}=\mathrm{K}_{\mathrm{a}}=10^{-8} \quad\text{..... (2)}\\ \Rightarrow & \frac{10^{-6} \alpha}{1-\alpha}=10^{-8} \end{aligned}$

Data given is inconsistent & contradictory. This should be bonus.

To calculate the percent degree of ionization of the compound $\mathrm{MX}_2$, we begin by considering its dissociation process:

$ \mathrm{MX}_2 \rightarrow \mathrm{M}^{+2} + 2\mathrm{X}^{-} $

Next, we use the van't Hoff factor (i), calculated as the ratio of the normal molar mass to the observed molar mass:

$ i = \frac{\text{normal molar mass}}{\text{observed molar mass}} = \frac{164}{65.6} $

Now, expressing the relationship of ionization in terms of the van't Hoff factor, we have:

$ 1 + (3 - 1)\alpha = \frac{164}{65.6} $

Simplifying this equation gives us:

$ 2\alpha = \frac{164}{65.6} - 1 = \frac{98.4}{65.6} $

Solving for $\alpha$:

$ \alpha = 0.75 $

Therefore, the percent dissociation, or the percent degree of ionization, is:

$ \text{Percent dissociation} = 75\% $

$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{C}_2 \mathrm{H}_2 \mathrm{NH}_3^{+}+\stackrel{\ominus}{\mathrm{O}} \mathrm{H}$

$\begin{array}{ccc} \mathrm{C}=10^{-3} \mathrm{M} & - & - \\ \mathrm{C}(1-\alpha) & \mathrm{C} \alpha & \mathrm{C} \alpha \\ \Rightarrow \mathrm{C}=10^{-3} & =10^{-5} & =10^{-5} \end{array}$

$1-\alpha \simeq 1$

Given, $\mathrm{P}^{\mathrm{H}}=9 \Rightarrow \mathrm{P}^{\mathrm{OH}}=5 \Rightarrow[\stackrel{\ominus}{\mathrm{O}} \mathrm{H}]=10^{-5} \mathrm{M}$

Now, $K_b=\frac{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_3^{+}\right][\stackrel{\ominus}{\mathrm{O}} \mathrm{H}]}{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\right]}$

$\Rightarrow \mathrm{K}_{\mathrm{b}}=\frac{10^{-5} \times 10^{-5}}{10^{-3}}=10^{-7}$

$\begin{aligned} & \mathrm{Ka}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \\ & 1.2 \times 10^{-5}=(0.03)\left(\alpha^2\right) \\ & \alpha=0.02 \\ & \pi=\mathrm{iCRT} \\ &=(1.02)(0.03)(0.083)(300) \\ &=0.76194 \\ &=76.194 \times 10^{-2} \mathrm{bar} \\ & \text { Nearest integer }=76 \end{aligned}$

To find the pH of an ammonium acetate solution, we use the fact that ammonium acetate is a salt resulting from the neutralization of a weak acid (acetic acid, $CH_3COOH$) by a weak base (ammonium hydroxide, $NH_4OH$). The respective ionization constant values for the weak acid $K_a$ and the weak base $K_b$ are given as $1.8 \times 10^{-5}$.

Firstly, calculate the $pK_a$ and $pK_b$ values.

• $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5})$
• $pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5})$

Given that the values of $K_a$ and $K_b$ are the same, their $pK_a$ and $pK_b$ values will also be the same, establishing a neutral condition where the effects of the acid and base neutralize each other.

The formula used to determine the pH of a solution of such a salt is:

$pH = \frac{1}{2}(pK_w + pK_a - pK_b)$

Where $pK_w$ is the ionic product of water, which is 14 at 25°C. In this specific case, since $pK_a = pK_b$, the formula simplifies to:

$pH = \frac{1}{2}(14 + pK_a - pK_a)$

$pH = \frac{1}{2}(14)$

$pH = 7$

The pH of an ammonium acetate solution in this scenario is 7, indicating a neutral solution.

$\begin{array}{ccc} & \text { 1M Benzoic acid } & +1 \mathrm{M} \text { Sodium Benzoate } \\ & \left(\mathrm{V}_{\mathrm{a}} \mathrm{ml}\right) & \left(\mathrm{V}_{\mathrm{s}} \mathrm{ml}\right) \\ \text { Millimole } & \mathrm{V}_{\mathrm{a}} \times 1 & \mathrm{~V}_{\mathrm{s}} \times 1 \end{array}$

$\begin{aligned} & \mathrm{pH}=4.5 \\ & \mathrm{pH}=\mathrm{pka}+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & 4.5=4.2+\log \left(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{a}}}\right) \\ & \frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{a}}}=2 \quad \text{..... (1)}\\ & \mathrm{~V}_{\mathrm{s}}+\mathrm{V}_{\mathrm{a}}=300 \quad \text{..... (2)}\\ & \mathrm{~V}_{\mathrm{a}}=100 \mathrm{~ml} \end{aligned}$

$\begin{aligned} & \text { Precipitation when } Q_{s p}=K_{s p} \\ & {\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2=10^{-11}} \\ & 0.1 \times\left[\mathrm{OH}^{-}\right]^2=10^{-11} \Rightarrow\left[\mathrm{OH}^{-}\right]=10^{-5} \\ & \Rightarrow \mathrm{pOH}=5 \quad \Rightarrow \mathrm{pH}=9 \end{aligned}$

First, we need to find the moles of NaOH and acetic acid (CH₃COOH) in the solution:

Moles of NaOH = Volume × Molarity = 20 mL × 0.1 M = 2 mmol

Moles of acetic acid = Volume × Molarity = 50 mL × 0.1 M = 5 mmol

Since NaOH is a strong base, it will react with acetic acid to form acetate ions (CH₃COO⁻) and water:

CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

2 mmol of NaOH will react with 2 mmol of acetic acid, resulting in 2 mmol of acetate ions and leaving 3 mmol of acetic acid unreacted.

Next, we need to find the concentrations of acetic acid and acetate ions in the resulting 70 mL solution:

Concentration of acetic acid = Moles / Total volume = 3 mmol / 70 mL = 0.04286 M

Concentration of acetate ions = Moles / Total volume = 2 mmol / 70 mL = 0.02857 M

Now we can use the Henderson-Hasselbalch equation to find the pH of the resulting solution:

pH = pKa + log ([A⁻] / [HA])

Given the pKa of acetic acid is 4.76, we can substitute the values:

pH = 4.76 + log (0.02857 / 0.04286)

Using the given log values, we can approximate the log value:

log (0.02857 / 0.04286) ≈ log (2/3) ≈ log 2 - log 3 ≈ 0.30 - 0.48 = -0.18

Now substitute this value back into the Henderson-Hasselbalch equation:

pH = 4.76 - 0.18 = 4.58

To express the pH as the nearest integer multiplied by 10⁻², multiply the pH value by 100:

4.58 × 100 = 458

So, the pH of the resulting solution is approximately 458 × 10⁻².

Initial concentrations before mixing:

$[\mathrm{Ba(NO_3)_2}] = 0.050\, \mathrm{M}$

$[\mathrm{NaF}] = 0.020\, \mathrm{M}$

Volumes of the solutions:

$V_{\mathrm{Ba(NO_3)_2}} = V_{\mathrm{NaF}} = 25.0\, \mathrm{mL}$

After mixing, the total volume becomes:

$V_{\mathrm{total}} = 25.0\, \mathrm{mL} + 25.0\, \mathrm{mL} = 50.0\, \mathrm{mL}$

Now, we calculate the initial concentrations after mixing:

$[\mathrm{Ba}^{2+}] = \frac{25.0\, \mathrm{mL} \times 0.050\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.025\, \mathrm{M}$

$[\mathrm{F}^{-}] = \frac{25.0\, \mathrm{mL} \times 0.020\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.010\, \mathrm{M}$

Next, we calculate the reaction quotient (Q) for the precipitation of BaF₂:

$Q = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$

$Q = (0.025\, \mathrm{M})(0.010\, \mathrm{M})^2 = 2.5 \times 10^{-6}$

K_sp of BaF₂ is given as $5 \times 10^{-7}$.

Now, we find the ratio of $[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$ to K_sp:

$\text{Ratio} = \frac{(2.5 \times 10^{-6})}{(5 \times 10^{-7})} = 5$

So, the correct ratio of $[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$ to K_sp is 5.

We can use the formula for pH to calculate the concentration of hydrogen ions in each solution:

$\mathrm{pH} = -\log_{10}[\mathrm{H}^+]$

For the first solution, we have:

$1 = -\log_{10}[\mathrm{H}^+]$

Solving for $[\mathrm{H}^+]$, we get:

$[\mathrm{H}^+] = 0.1 \mathrm{~M}$

For the second solution, we want:

$2 = -\log_{10}[\mathrm{H}^+]$

Solving for $[\mathrm{H}^+]$, we get:

$[\mathrm{H}^+] = 0.01 \mathrm{~M}$

To dilute the first solution to the desired concentration, we can use the dilution equation:

$C_1V_1 = C_2V_2$

where $C_1$ and $V_1$ are the initial concentration and volume, and $C_2$ and $V_2$ are the final concentration and volume.

We know $C_1 = 0.1 \mathrm{~M}$, $C_2 = 0.01 \mathrm{~M}$, and $V_1 = 1 \mathrm{~L}$. Solving for $V_2$, we get:

$V_2 = \frac{C_1V_1}{C_2} = \frac{(0.1 \mathrm{~M})(1 \mathrm{~L})}{0.01 \mathrm{~M}} = 10 \mathrm{~L}$

Therefore, we need to add $10-1=9$ liters of water to the initial solution to obtain the desired pH. Converting liters to milliliters, we get:

$9 \mathrm{~L} \times \frac{1000 \mathrm{~mL}}{1 \mathrm{~L}} = 9000 \mathrm{~mL}$

So the volume of water needed is 9000 mL or 9,000 mL (nearest integer).

Barium sulfate, BaSO₄, is a sparingly soluble salt. Its dissolution can be represented by the following equilibrium reaction:

$\mathrm{BaSO}_4 (\mathrm{s}) \rightleftharpoons \mathrm{Ba}^{2+}(\mathrm{aq}) + \mathrm{SO}_4^{2-}(\mathrm{aq})$

The solubility product constant, Ksp, is given by:

$\mathrm{K}_{sp} = [\mathrm{Ba}^{2+}][\mathrm{SO}_4^{2-}]$

In this case, the salt is being dissolved in a solution that already contains sulfate ions, $\mathrm{SO}_4^{2-}$, from the $\mathrm{K}_2\mathrm{SO}_4$.

When $\mathrm{K}_2\mathrm{SO}_4$ dissolves completely in water, it forms $\mathrm{K}^+$ and $\mathrm{SO}_4^{2-}$ ions. Because its concentration is 0.1 M, the $\mathrm{SO}_4^{2-}$ concentration due to $\mathrm{K}_2\mathrm{SO}_4$ is 0.1 M.

Therefore, the concentration of $\mathrm{SO}_4^{2-}$ ions is now not just due to the BaSO₄ dissolving, but also the added $\mathrm{K}_2\mathrm{SO}_4$. Hence, the total concentration of $\mathrm{SO}_4^{2-}$ ions is $\mathrm{S} + 0.1$ where S is the solubility of $\mathrm{BaSO}_4$.

We then substitute into the Ksp expression:

$1 \times 10^{-10} = [\mathrm{S}][\mathrm{S} + 0.1]$

However, because BaSO₄ is sparingly soluble, S is very small compared to 0.1. Therefore, we can make the approximation that $\mathrm{S} + 0.1$ is approximately 0.1. This simplifies the equation to:

$1 \times 10^{-10} = 0.1[\mathrm{S}]$

Solving for S (the molar solubility of BaSO₄) gives:

$S = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \, \mathrm{mol/L}$

To convert this molarity to a mass per volume concentration, we multiply by the molar mass of BaSO₄, which is 233 g/mol:

$\mathrm{Solubility} = S \times \text{Molar mass of BaSO4} = 1 \times 10^{-9} \, \mathrm{mol/L} \times 233 \, \mathrm{g/mol} = 233 \times 10^{-9} \, \mathrm{g/L}$

So, the solubility of BaSO₄ in a 0.1 M K₂SO₄ solution is 233 x $10^{-9}$ g/L, or 233 ng/L.

A. This statement is incorrect. Phenolphthalein cannot be used as an indicator for the titration of a weak acid with a weak base because the pH at the equivalence point is not within the color change range of phenolphthalein (8.2-10.0).

B. This statement is correct. Phenolphthalein begins to change color at a pH of around 8.2, and the transition completes around pH 10.0. It transitions from colorless to pink as the pH increases past this point.

C. This statement is incorrect. Phenolphthalein is not a base; it is a pH indicator, which can donate protons (like an acid) but not accept them (like a base).

D. This statement is correct. Phenolphthalein is indeed colorless in acidic medium (pH less than 7), and it turns pink in a basic solution (pH greater than 7).

So, the number of correct statements is 2.

$ \begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left(\mathrm{Ag}^{+}\right)\left(\mathrm{Cl}^{-}\right)=\mathrm{S} \times \mathrm{S}=\mathrm{S}^2 \\\\ & \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}} \\\\ & \mathrm{S}=\frac{1.434 \times 10^{-3}}{143.4}=10^{-5} \\\\ & \mathrm{~K}_{\mathrm{sp}}=\mathrm{S}^2=10^{-10} \\\\ & \Rightarrow-\log \left(\mathrm{K}_{\mathrm{sp}}\right)=-\log \left(10^{-10}\right)=10 \end{aligned} $

$\mathrm{[H^+]=\frac{6+8}{1000}=14\times10^{-3}}$

$\mathrm{pH=3-\log14}$

$=3-.3-.84$

$=1.86=186\times10^{-2}$

$ \begin{aligned} & \mathrm{pH}=12, \\\\ & \mathrm{pH}+\mathrm{pOH}=14 \\\\ & \mathrm{pOH}=14-12=2 \\\\ & {\left[\mathrm{OH}^{-}\right]=10^{-2} \mathrm{~mol} \mathrm{~L}} \\\\ & \underset{5 \times 10^{-3}}{\mathrm{Ca}(\mathrm{OH})_2} \longrightarrow \underset{5 \times 10^{-3}}{\mathrm{Ca}^{2+}}+\underset{10^{-2}}{2 \mathrm{OH}^{-}} \end{aligned} $

Moles of $\mathrm{Ca}(\mathrm{OH})_2$ in $1000 \mathrm{~mL}=5 \times 10^{-3}$

Millimoles in $100 \mathrm{~mL}=5 \times 10^{-1}$

In resultant solution

$ \begin{aligned} & \mathrm{n}_{\mathrm{NH}_3}= 0.1-0.02=0.08 \\\\ & \mathrm{n}_{\mathrm{NH}_4 \mathrm{Cl}}= \mathrm{n}_{\mathrm{NH}_4^{+}}=0.1+0.02=0.12 \\\\ & \mathrm{pOH}= \mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\\\ &=4.745+\log \frac{0.12}{0.08} \\\\ &=4.745+\log \frac{3}{2} \\\\ &=4.745+0.477-0.301 \\\\ & \mathrm{pOH}=4.921 \\\\ & \mathrm{pH}=14-\mathrm{pH} \\\\ &=9.079 = 9079\times 10^{-3} \end{aligned} $

Concentration of calcium lactate $=0.005 \mathrm{M}$, concentration of lactate ion $=(2 \times 0.005) \mathrm{M}$.

Calcium lactate is a salt of weak acid $+$ strong base

$\therefore$ Salt hydrolysis will take place.

$ \begin{aligned} & \mathrm{pH}=7+\frac{1}{2}(\mathrm{pKa}+\log \mathrm{C}) \\\\ & =7+\frac{1}{2}(5+\log (2 \times 0.005)) \\\\ & =7+\frac{1}{2}[5-2 \log 10]=7+\frac{1}{2} \times 3=8.5=85 \times 10^{-1} \end{aligned} $

To find the dissociation constant of acetic acid, we use the Henderson-Hasselbalch equation for the given system and conditions. The equation is as follows:

$ \text{pH} = \text{pK}_\text{a} + \log \left(\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}\right) $

In the solution mixture:

We have 25 mL of 0.2 M $\text{CH}_3\text{COONa}$ and 25 mL of 0.02 M $\text{CH}_3\text{COOH}$.

The concentration ratio $\left(\frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]}\right)$ becomes $\frac{25 \times 0.2}{25 \times 0.02} = 10$.

Given that the pH of the solution is 5, we substitute into the equation:

$ 5 = \text{pK}_\text{a} + \log 10 $

Since $\log 10 = 1$, we solve for $\text{pK}_\text{a}$:

$ 5 = \text{pK}_\text{a} + 1 \quad \Rightarrow \quad \text{pK}_\text{a} = 4 $

Converting from $\text{pK}_\text{a}$ to $K_\text{a}$, we use:

$ K_\text{a} = 10^{-\text{pK}_\text{a}} = 10^{-4} $

Thus, since the dissociation constant $K_\text{a}$ is given as $x \times 10^{-5}$, compare:

$ 10^{-4} = 10 \times 10^{-5} $

Therefore, $x = 10$.

$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}$

$\mathrm{S}=\sqrt{K_{s p}}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$

$=2.82 \times 10^{-14}$

$=282 \times 10^{-16}$

$ \therefore $ Ans. 282

$\mathrm{K}_{\mathrm{a}}$ of Butyric acid $\Rightarrow 2 \times 10^{-5} \,\mathrm{PKa}=4.7$

$\mathrm{pH}$ of $0.2 \mathrm{M}$ solution,

$ \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{C} $

$ \begin{aligned} &=\frac{1}{2}(4 \cdot 7) - \frac{1}{2} \log (0.2) \\\\ &=2.35+0.35=2.7 \end{aligned} $

$ \mathrm{pH}=27 \times 10^{-1} $

$\mathrm{CaF}_{2} \stackrel{\mathrm{s}}{\rightleftharpoons} \underset{ \mathrm{s}}{\mathrm{Ca}^{2+}}+\underset{2 \mathrm{~s}}{2 \mathrm{~F}^{-}}$

$ \begin{aligned} \mathrm{K}_{\mathrm{sp}} &=\mathrm{s}(2 \mathrm{~s})^{2} \\\\ &=4 \mathrm{~s}^{3} \end{aligned} $

Solubility $(\mathrm{s})=2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$

$=\frac{2 \cdot 34 \times 10^{-3} \times 10}{78}$ mole $/$ lit

$=3 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$

$\therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(3 \times 10^{-4}\right)^{3}$

$ \begin{aligned} &=108 \times 10^{-12} \\\\ &=0.0108 \times 10^{-8}(\mathrm{~mole} / \mathrm{lit})^{3} \end{aligned} $

$ \begin{aligned} & \therefore x \approx 0 \end{aligned} $

$16 \mathrm{H}^{+}+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 10 \mathrm{CO}_{2}+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}$

During titration of oxalic acid by $\mathrm{KMnO}_{4}$, oxalic acid converts into $\mathrm{CO}_{2}$.

$\therefore$ Change in oxidation state of carbon $=1$

$A_{2} X_{3} \rightleftharpoons \underset{2S}{2 \mathrm{~A}}+\underset{3S}{3 \mathrm{X}}$

$ \begin{aligned} &\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^{2}(3 s)^{3}=1.1 \times 10^{-23} \\\\ &\mathrm{~S} \approx 10^{-5} \end{aligned} $

For sparingly soluble salts

$ \begin{aligned} \wedge_{m} &=\wedge_{m}^{0} \\\\ \wedge_{m} &=\frac{\mathrm{k}}{\mathrm{S} \times 10^{3}} \\\\ &=\frac{3 \times 10^{-5}}{10^{-5}} \times 10^{-3} \\\\ &=3 \times 10^{-3} ~ \mathrm{Sm}^{2} \mathrm{~mol}^{-1} \end{aligned} $

$ \begin{aligned} &{\left[\mathrm{OH}^{-}\right]=0.001=10^{-3} \mathrm{M}} \\\\ &{\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}} \\\\ &\quad\left[\mathrm{H}^{+}\right]=10^{-11} \\\\ &\begin{aligned} \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\\\ =&-\log \left(10^{-11}\right) \\\\ \mathrm{pH} &=11 \end{aligned} \end{aligned} $

CH₃COOH + NaOH $\to$ CH₃COONa + H₂O

After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml

Initially,

Number of millimole of NaOH = 25 $\times$ 0.1 = 2.5 mm

Number of millimole of CH₃COOH = 50 $\times$ 0.1 = 5 mm

After nutrilisation,

Millimole of NaOH = 0

Millimole of CH₃COOH = 5 $-$ 2.5 = 2.5 mm

Millimole of CH₃COONa = 2.5

After nutrilisation,

Concentration of CH₃COOH = $[C{H_3}COOH] = {{5 - 2.5} \over {75}} = {1 \over {30}}$

Concentration of CH₃COONa = $[C{H_3}COONa] = {{ 2.5} \over {75}} = {1 \over {30}}$

${P^H} = {P^{Ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$

$ = 4.76 + \log {{{1 \over {30}}} \over {{1 \over {30}}}}$

$ = 4.76 + \log (1)$

$ = 4.76 + 0$

$ = 4.76$

$ = 4.76 \times {10^{ - 2}}$

Due to common-ion effect (presence of NaOH) the concentration of OH^$-$ will be (2S + 0.1) $\approx$ 0.1

($\because$ 0.1 > > 2 S)

$\therefore$ Solubility of product,

${K_{sp}} = {(0.1)^2} \times S$

$2 \times {10^{ - 20}} = 0.01 \times S$

$ \Rightarrow S = {{2 \times {{10}^{ - 20}}} \over {0.01}} = 2 \times {10^{ - 18}}$

$\therefore$ x = 2

$[HCl] = {{20} \over {80}} = {1 \over 4}M = 2.5 \times {10^{ - 1}}M$

pH = $-$log 2.15 $\times$ 10^-1 = 1 $-$ 0.3979 = 0.6021

pH = 6021 $\times$ 10^-4

K_SP = (3s)³ (2s)²

K_SP = 108 s⁵ & s = (x/M)

K_SP = 108${\left( {{x \over M}} \right)^5}$

given ${K_{sp}} = a{\left( {{x \over M}} \right)^5}$

comparing a = 108

In pure water



$ \therefore $ K_sp = (s)² = (8 $\times$ 10^-4)² = 64 $\times$ 10^-8

In H₂SO₄ solution,



As S₁ < < 0.01 so, S₁ + 0.01 $ \simeq $ 0.01

$ \therefore $ K_sp = [Cd⁺²] [So$_4^{ - 2}$]

$ \Rightarrow $ 64 $\times$ 10^-8 = S₁ $\times$ 0.01

$ \Rightarrow $ S₁ = 64 $\times$ 10^-6

$C{H_3}COOH + C{H_3}COONa$ [Acidic Buffer]

${pH} = {p{ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$

$ \Rightarrow 5.74 = 4.74 + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$

$ \Rightarrow 1 = \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$

$ \Rightarrow {{[C{H_3}COONa]} \over {[C{H_3}COOH]}} = 10$

$ \Rightarrow [C{H_3}COONa] = 10 \times 1 = 10$

$K{a_1}$ of ${H_2}S{O_3} > > K{a_2}$ of ${H_2}S{O_3}$

$ \therefore $ The contribution of H⁺ from 2^nd dissociation of H₂SO₃ can be neglected.



$ \Rightarrow $ ${{c{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$

$ \Rightarrow {{0.588{\alpha ^2}} \over {1 - \alpha }} = 1.7 \times {10^{ - 2}}$

$ \Rightarrow 58.8{\alpha ^2} = 1.7 - 1.7\alpha $

$ \Rightarrow 58.8{\alpha ^2} + 1.7\alpha - 1.7 = 0$

$\alpha = {{ - 1.7 + \sqrt {{{1.7}^2} + 4 \times 1.7 \times 58.8} } \over {2 \times 58.8}} = 0.156$

$[{H^ + }] = c\alpha = 0.092$

$pH = - \log [{H^ + }]$

$ = 1.036$

$ \approx 1$

For A₂X

$\matrix{ {{A_2}X} & \to & {2{A^ + }} & {{X^{2 - }}} \cr {} & {} & {2{S_1}} & {{S_1}} \cr } $

${K_{sp}} = 4S_1^3 = 4 \times {10^{ - 12}}$

S₁ = 10^$-$4

for MX

$\matrix{ {MX} & \to & {{M^ + }} & {{X^ - }} \cr {} & {} & {{S_2}} & {{S_2}} \cr } $

${K_{sp}} = S_2^2 = 4 \times {10^{ - 12}}$

S₂ = 2 $\times$ 10^$-$6

so, ${{{S_{{A_2}X}}} \over {{S_{MX}}}} = {{{{10}^{ - 4}}} \over {2 \times {{10}^{ - 6}}}} = 50$

Since (NH₄)₃PO₄ is salt of weak acid (H₃PO₄) & weak base (NH₄OH).

pH = 7 + ${1 \over 2}$(pk_a $-$ pk_b)

= 7 + ${1 \over 2}$ (5.23 $-$ 4.75)

= 7.24 $ \approx $ 7.

Given, ${[{K_{sp}}]_{Pb{l_2}}} = 8 \times {10^{ - 9}}$

To calculate solubility of Pbl₂ in 0.1 M solution of Pb(NO₃)₂,

(I) $\mathop {Pb{{(N{O_3})}_2}}\limits_{0. 1 M} \to \mathop {P{b^{2 + }}(aq)}\limits_{0.1 M} + \mathop {2NO_3^ - (aq)}\limits_{0. 2 M} $

(II) $Pb{I_2}(s)$ $\rightleftharpoons$ $\mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2{I^ - }(aq)}\limits_{2S} $

$\therefore$ [Pb²⁺] = S + 0.1 $\approx$ 0.1

$\because$ S < < 0.1

Now, K_sp = 8 $\times$ 10^$-$9

[Pb²] [I^$-$]² = 8 $\times$ 10^$-$9

0.1 $\times$ (2S)² = 8 $\times$ 10^$-$9

4S² = 8 $\times$ 10^$-$8 $\Rightarrow$ S = 141 $\times$ 10^$-$6 M

$ \therefore $ x = 141

AB2	⇌	A2+(aq)	+	2B-(aq)
		s		2s

K_sp = 4s³ = 3.2 × 10^–11

$ \Rightarrow $ s³ = 8 × 10^–12

$ \Rightarrow $ s = 2 × 10^–4

CO₂ + H₂O $ \to $ H₂CO₃

At 30 bar pressure mass of CO₂ in 1 kg water = 44 gm

At 3 bar pressure mass of CO₂ in 1 kg water = 4.4 gm

$ \therefore $ Moles of CO₂ in 1 kg water = ${{4.4} \over {44}}$ = 0.1

	H2CO3	⇌	H+	+	HCO3-
t = 0	0.1		0		0
t = teq	0.1(1 - $\alpha $)		0.1$\alpha $		0.1$\alpha $

4.0 $ \times $ 10^–7 = ${{0.1{\alpha ^2}} \over {1 - \alpha }}$

${1 - \alpha }$ $ \simeq $ 1

$ \Rightarrow $ 0.1${{\alpha ^2}}$ = 4 $ \times $ 10^-7

$ \Rightarrow $ $\alpha $ = 2 $ \times $ 10^-3

[H⁺] = 0.1$\alpha $ = 2 $ \times $ 10^-4

$ \therefore $ pH = –[– 4 × log(2)] = 3.7 = 37 × 10^–1

milimole of acetic acid in 20 ml
= ${3 \over {60}} \times 1000 \times {{20} \over {500}}$ = 2

milimole of HCl in 20 ml = 25 $ \times $ ${{20} \over {500}}$ = 1

milimole of NaOH 20 ml = ${1 \over 2} \times 5$ = 2.5

NaOH	+	CH3COOH	$ \to $	CH3COONa	+	H2O
	1.5	2		0		0
	0	0.5		1.5

pH = pKa + log${{1.5} \over {0.5}}$

= 4.74 + log 3 = 5.22

In 100 L solution H₂SO₄ present = 9.8 gm

$ \therefore $ In 10 L solution H₂SO₄ present = $9.8 \times {{10} \over {100}}$ gm

$ \Rightarrow $ In 10 L solution moles of H₂SO₄ present = ${{9.8} \over {98}} \times {{10} \over {100}}$

In one molecule of H₂SO₄ two H⁺ ion present.

$ \therefore $ In 10 L solution moles of H⁺ present = 2$ \times $${{9.8} \over {98}} \times {{10} \over {100}}$ = 0.02 moles

Also In 100 L solution NaOH present = 4 gm

$ \therefore $ In 40 L solution NaOH present = $4 \times {{40} \over {100}}$

$ \Rightarrow $ In 40 L solution moles of NaOH present = ${4 \over {40}} \times {{40} \over {100}}$

In one molecule of NaOH one OH^- ion present.

$ \therefore $ In 40 L solution moles of OH^- ion present = ${4 \over {40}} \times {{40} \over {100}}$ = 0.04 moles

As moles of OH^- ion is more than H⁺ ion, so solution is basic.

$ \therefore $ Final Conc. of OH^– = ${{0.04 - 0.02} \over {40 + 10}}$ = 4 $ \times $ 10^-4

$ \therefore $ pOH = – log (4 ×10^–4) = 3.4

pH = 14 – 3.4 = 10.6

Calculate the concentration of iodate ions $\left[\text{IO}_3^-\right]$ in the equilibrium mixture:

The concentration in the prepared solution is given by:

$ \left[\text{IO}_3^-\right]_{\text{eqm}} = \frac{6}{300} = 0.02 \, \text{M} $

Write the expression for the solubility product constant ($ K_{\text{sp}} $) for barium iodate:

$ K_{\text{sp}} = [\text{Ba}^{2+}][\text{IO}_3^-]^2 $

Rearrange to solve for the concentration of barium ions $[\text{Ba}^{2+}]$:

$ [\text{Ba}^{2+}] = \frac{1.58 \times 10^{-9}}{(0.02)^2} = 3.95 \times 10^{-6} \, \text{M} $

The solubility of $\text{Ba}(\text{IO}_3)_2$, which is the same as the concentration of barium ions in this context, is:

$ X = 3.95 $

Thus, the solubility constant $ X $ is $ 3.95 $, and the solubility of barium iodate is $ 3.95 \times 10^{-6} \, \text{mol dm}^{-3} $.

$ \text { Temperature }=25^{\circ} \mathrm{C} $Concentration of aqueous solution of weak monobasic acid $=1.00 \times 10^{-3} \mathrm{~m}$Arid dissociation constant $\left(K_a\right)$ of the weak monobasic acid $=4.00 \times 10^{-11}$$ \text { Ionic product of water }\left(K_w\right)=1.00 \times 10^{-14} $Concentration of $\mathrm{H}^{+}$ions $=X \times 10^{-7} \mathrm{M}$ Consider$ H A \rightleftharpoons H^{+}+A^{-}HA - \,\,\,\,\,\,\,monobasic \,\,Weak \,\,\,acid $$ C_1 $$ c_1\left(1-\alpha_1\right) \quad c_1 \alpha_1 \quad c_1 \alpha_1 $$ \text { So, } K_a=\frac{\left[H^{+}\right][A^-]}{[H A]} $$ =\frac{\left[H^{+}\right] c_1 \alpha_1}{c_1\left(1-\alpha_1\right)} \quad \alpha_1 \ll 1 \quad \text { so, } 1-\alpha_1 \approx 1 $$ =\frac{\left[H^{+}\right] c_1 \alpha_1}{c_1}, k a c_1=\left[H^{+}\right] c_1 \alpha_1-(1) $$ \begin{aligned} &\text { Consider, }\\ &\begin{aligned} & \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \\ & \mathrm{C}_2\left(1-\alpha_2\right) \mathrm{C}_2 \alpha_2 \quad \mathrm{C}_2 \alpha_2 \end{aligned} \end{aligned} $So, $ \begin{aligned} K_W & =\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ K_W & =\left[\mathrm{H}^{+}\right] \mathrm{C}_2 \alpha_2 \,\,\,\,\,\,\,...(2) \end{aligned} $ Add (1) and (2) $\Rightarrow$$ K_a C_1+K_w=\left[H^{+}\right] C_1 \alpha_1+\left[H^{+}\right] C_2 \alpha_2=\left[H^{+}\right]\left(C_1 \alpha_1+C_2 \alpha_2\right) $ Concentration of $\mathrm{H}^{+}$has contribution from, both HA and $\mathrm{H}_2 \mathrm{O}$. So, $\left[\mathrm{H}^{+}\right]=c_1 \alpha_1+\mathrm{C}_2 \alpha_2$ Substitute for $\left[H^A\right]$ as,$ K_a c_1+K_w=\left[H^{+}\right]\left(c_1 \alpha_1+c_2 \alpha_2\right) $$ \begin{aligned} K_a C_1+K_w & =\left[\mathrm{H}^{+}\right]\left[\mathrm{H}^{+}\right] \\ & =\left[\mathrm{H}^{+}\right]^2 \end{aligned} $$ \begin{aligned} &\text { So, }\\ &\begin{aligned} {\left[H^{+}\right] } & =\sqrt{K_a C_1+K w} \\ & =\sqrt{4.00 \times 10^{-11} \times 1.00 \times 10^{-3}+1.00 \times 10^{-14}} \\ & =\sqrt{4.00 \times 10^{-14}+1.00 \times 10^{-14}} \\ & =\sqrt{5.00 \times 10^{-14}} \\ & =2.24 \times 10^{-7} \mathrm{M}\left(\times \times 10^{-7}\right) \end{aligned} \end{aligned} $So, value of $X$ is 2.24 Answer: 2.24

Given,

Conentration of $\mathrm{H}_2 \mathrm{SO}_4=1 \mathrm{M}$

Conentration of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ = $1.8 \times 10^{-2} \mathrm{M}$



From the above two equations, we get

$ \left[\mathrm{H}^{+}\right]=1 \mathrm{M} \text { and }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2} $



$ K_c=\frac{1.8 \times 10^{-2} \times 1}{1}=1.8 \times 10^{-2} $

and it is given that $\mathrm{K}_{a_2}\left(\mathrm{Q}_c\right)=1.2 \times 10^{-2} \mathrm{M}$

Since, $\mathrm{K}_{a_2}$ (i.e., $\mathrm{Q}_c$ ) $<\mathrm{K}_c$

$\therefore$ Rather than dissociation of $\mathrm{HSO}_4^{-}$into $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ ions, association between already present $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ will take place.

Assuming ' $x$ ' mol/L of $\mathrm{SO}_4^{2-}$ and $\mathrm{H}^{+}$combines to form $\mathrm{HSO}_4^{-}$



$\begin{aligned} & \mathrm{K}_{a 2}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{HSO}_4^{-}\right]} \\\\ & 1.2 \times 10^{-2}=\frac{(1-x)\left(1.8 \times 10^{-2}-x\right)}{(1+x)} \\\\ & \because x<<1, \text { so }(1+x) \approx 1 \text { and }(1-x) \approx 1 \\\\ & 1.2 \times 10^{-2}=1.8 \times 10^{-2}-x \\\\ & x=\left(1.8 \times 10^{-2}\right)-\left(1.2 \times 10^{-2}\right) \\\\ & x=0.6 \times 10^{-2} \mathrm{M} \\\\ & \text { So, }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2}-x \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=\left(1.8 \times 10^{-2}\right)-\left(0.6 \times 10^{-2}\right) \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=1.2 \times 10^{-2} \mathrm{M}\end{aligned}$



Given, $\quad \mathrm{K}_{s p}=1.6 \times 10^{-8}$

$ \therefore $ $y\left(1.2 \times 10^{-2}+y\right)=1.6 \times 10^{-8}$

Since, $y<<1$, So $1.2 \times 10^{-2}+y \approx 1.2 \times 10^{-2}$

So, $y \times 1.2 \times 10^{-2}=1.6 \times 10^{-8}$

$ \Rightarrow $ $y =\frac{1.6 \times 10^{-8}}{1.2 \times 10^{-2}}$

$ \Rightarrow $ $ y =1.33 \times 10^{-6} $

$ X \times 10^{-Y} \mathrm{M} =1.33 \times 10^{-6} \mathrm{M} $

So, $\mathrm{Y}=6$

Hence, the value of $Y$ is 6 .

First acid base reaction between H₂CO₃ and NaOH takes place.

In the final solution, we have 0.01 mole Na₂CO₃ and 0.02 moles of NaHCO₃.

Here, we have a buffer of NaHCO₃ and Na₂CO₃.

$\therefore$ $pH = p{K_{{a_2}}} + \log {{[Salt]} \over {[Acid]}}$

$ = 10.32 + \log {{\left( {{{0.01} \over {0.1}}} \right)} \over {\left( {{{0.02} \over {0.1}}} \right)}}$

$ = 10.32 + \log {1 \over 2}$

$ = 10.32 - \log 2$

$ = 10.32 - 0.3$

$ = 10.02$

$\therefore$ $pH = 10.02$

From the given diagram, 63 mL volume of HA used till equivalence point. At half of equivalence point, solution will be basic buffer with B and BH⁺.

$pOH = p{K_b} + \log {{[B{H^ + }]} \over {[B]}}$

At half equivalence point :

[BH⁺] = [B] ($ \because $ pH = 11)

Therefore, pOH = pK_b = 14 $-$ 11 = 3

$ \because $ pK_b = 3.00