Ionic Equilibrium

For a buffer of weak base $B$ and its conjugate acid $BH^+$, we use:

$\mathrm{pH}=\mathrm{p}K_a+\log\left(\frac{[B]}{[BH^+]}\right)$

Step 1: Find $\mathrm{p}K_a$ of $BH^+$

Given $\mathrm{p}K_b=5.699$.

At $25^\circ\mathrm{C}$, $\mathrm{p}K_w=14$ so

$\mathrm{p}K_a=14-\mathrm{p}K_b=14-5.699=8.301$

Step 2: Use buffer equation for pH $=9$

$9=8.301+\log\left(\frac{[B]}{[BH^+]}\right)$

$\log\left(\frac{[B]}{[BH^+]}\right)=9-8.301=0.699$

Given $\log 5=0.699$, so

$\frac{[B]}{[BH^+]}=5$

Step 3: Write moles after mixing HCl and base

Moles of HCl added:

$n(\mathrm{HCl})=0.02\cdot \frac{x}{1000}$

Moles of base $B$ taken:

$n(B)=0.02\cdot \frac{y}{1000}$

Reaction: $B+\mathrm{HCl}\rightarrow BH^+ + Cl^-$

So after reaction:

• $n(BH^+)=n(\mathrm{HCl})=0.02\cdot \frac{x}{1000}$

• $n(B)=n(B)_{\text{initial}}-n(\mathrm{HCl})=0.02\cdot\frac{(y-x)}{1000}$

Hence,

$\frac{[B]}{[BH^+]}=\frac{y-x}{x}=5$

So,

$y-x=5x \Rightarrow y=6x$

Step 4: Use total volume condition

Given final volume is $100\ \mathrm{mL}$:

$x+y=100$

Substitute $y=6x$:

$x+6x=100 \Rightarrow 7x=100 \Rightarrow x=14.2857$

$y=100-x=85.7143$

Answer

$x=14.3\ \mathrm{mL},\quad y=85.7\ \mathrm{mL}$

So, Option B.

For a basic buffer of $\mathrm{NH_4OH}$ and its salt $\mathrm{NH_4Cl}$, NCERT uses:

$\mathrm{pOH}=\mathrm{p}K_b+\log\left(\frac{[\text{salt}]}{[\text{base}]}\right)$

Given $\mathrm{p}K_b(\mathrm{NH_4OH})=4.75$ and required $\mathrm{pH}=9.25$:

$\mathrm{pOH}=14-9.25=4.75$

So,

$4.75=4.75+\log\left(\frac{[\text{salt}]}{[\text{base}]}\right)$

$\Rightarrow \log\left(\frac{[\text{salt}]}{[\text{base}]}\right)=0 \Rightarrow \frac{[\text{salt}]}{[\text{base}]}=1$

So, we need equal amounts of salt and base after mixing.

Reaction:

$\mathrm{NH_4OH}+\mathrm{HCl}\rightarrow \mathrm{NH_4Cl}+\mathrm{H_2O}$

If initial moles are $n_b$ (base) and $n_a$ (acid), then after reaction (for buffer, base must be in excess):

• moles of salt formed $=n_a$

• moles of base left $=n_b-n_a$

Condition for $\dfrac{[\text{salt}]}{[\text{base}]}=1$ is:

$n_a=n_b-n_a \Rightarrow n_b=2n_a$

Now check options by moles:

Option B

$n_b=0.2\times 0.5=0.10\ \text{mol}$

$n_a=0.1\times 0.5=0.05\ \text{mol}$

Here,

$n_b=0.10=2(0.05)=2n_a$

So, after reaction:

• salt $=0.05$ mol

• base left $=0.10-0.05=0.05$ mol

Equal $\Rightarrow$ required buffer, hence $\mathrm{pOH}=\mathrm{p}K_b=4.75$ and $\mathrm{pH}=9.25$.

Correct option: B

An aqueous solution of HCl initially has a pH of 1.0, meaning the hydrogen ion concentration is $[\text{H}^+]=10^{-1} \, \text{M}$.

When we dilute the solution by adding an equal volume of water, the concentration of hydrogen ions will be halved:

$ [\text{H}^+]_{\text{new}} = \frac{10^{-1}}{2} = 5 \times 10^{-2} \, \text{M} $

To find the new pH, we use the pH formula:

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

Substituting the new hydrogen ion concentration gives:

$ \text{pH} = -\log(5 \times 10^{-2}) $

Using the logarithm property $\log(a \times b) = \log a + \log b$:

$ \text{pH} = -(\log 5 + \log 10^{-2}) = -\log 5 + 2 $

Given $\log 2 = 0.30$, we need $\log 5$, which can be calculated as:

$ \log 10 = \log(2 \times 5) = \log 2 + \log 5 \Rightarrow \log 5 = \log 10 - \log 2 = 1 - 0.30 = 0.70 $

Thus:

$ \text{pH} = -(0.70) + 2 = 1.3 $

Therefore, the pH of the solution increases to 1.3.

For HCl:

$ \text{Moles of HCl} = \text{Moles of NaOH used} $

$ M \times 40 = 0.1 \times 2 $

$ M = 0.005 $

For $\mathrm{CH}_3 \mathrm{COOH}$:

$ \text{Moles of } \mathrm{CH}_3 \mathrm{COOH} = \text{Moles of NaOH used} $

$ M \times 40 = 0.1 \times 3 $

$ M = 0.0075 $

Since HCl is a strong acid, it is neutralized first.

When equal volumes of solutions are mixed, the molarity of each solution halves. To determine if a precipitate will form, calculate the ionic product $\mathrm{Q}_{\mathrm{SP}}$ using the formula:

$ \mathrm{Q}_{\mathrm{SP}} = \left[\mathrm{A}^{2+}\right]\left[\mathrm{Y}^{-}\right]^2 $

A precipitate of $\mathrm{AY}_2$ will form if $\mathrm{Q}_{\mathrm{SP}} > \mathrm{K}_{\mathrm{SP}}$, where $\mathrm{K}_{\mathrm{SP}}$ is given as $5.2 \times 10^{-7}$.

Option 1 Calculation:

$ \mathrm{Q}_{\mathrm{SP}} = \left(1.8 \times 10^{-3}\right)\left(\frac{5}{2} \times 10^{-4}\right)^2 = \text{Calculated value} < \mathrm{K}_{\mathrm{SP}} $

Option 2 Calculation:

$ \mathrm{Q}_{\mathrm{SP}} = \left(10^{-4}\right)\left(0.4 \times 10^{-3}\right)^2 = \text{Calculated value} < \mathrm{K}_{\mathrm{SP}} $

Option 3 Calculation:

$ \mathrm{Q}_{\mathrm{SP}} = \left(10^{-2}\right)\left(10^{-2}\right)^2 = \text{Calculated value} > \mathrm{K}_{\mathrm{SP}} $

Option 4 Calculation:

$ \mathrm{Q}_{\mathrm{SP}} = \left(\frac{1.5}{2} \times 10^{-4}\right)\left(\frac{1.5}{2} \times 10^{-3}\right)^2 = \text{Calculated value} < \mathrm{K}_{\mathrm{SP}} $

From the calculations, the solution in Option 3 will form a precipitate since the ionic product $\mathrm{Q}_{\mathrm{SP}}$ is greater than the solubility product $\mathrm{K}_{\mathrm{SP}}$.

Based on the Ksp values and salt analysis cation identification, we can say that order of Ksp value is:

$\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_2$

Ksp values

$\begin{aligned} & \mathrm{HgS} \rightarrow 4 \times 10^{-53} \\ & \mathrm{PbS} \rightarrow 8 \times 10^{-28} \\ & \mathrm{AgBr} \rightarrow 5 \times 10^{-13} \\ & \mathrm{Ca}(\mathrm{OH})_2 \rightarrow 5.5 \times 10^{-6} \end{aligned}$

$\begin{aligned} & \mathrm{HA} \rightleftharpoons \mathrm{H}^{\oplus}+\mathrm{A}^{\Theta} \\ & \mathrm{t}=0 \quad \mathrm{a} \\ & \mathrm{t}=\mathrm{t} \quad \mathrm{a}(1-\mathrm{x}) \quad \mathrm{ax} \quad \mathrm{ax} \\ & \mathrm{~K}_{\mathrm{a}}=(\mathrm{ax}) \frac{(\mathrm{x})}{1-\mathrm{x}} ;\left[\mathrm{H}^{+}\right]=\mathrm{ax} \\ & -\log \left(\mathrm{K}_{\mathrm{a}}\right)=-\log (\mathrm{ax})-\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right) \\ & \mathrm{pKa}=\mathrm{pH}-\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right) \\ & \mathrm{pH}-\mathrm{pKa}=\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right) \end{aligned}$

To find the molar solubility of $\mathrm{Cr}(\mathrm{OH})_3$ in water, we start with the dissolution equilibrium:

$ \mathrm{Cr}(\mathrm{OH})_{3(\mathrm{~s})} \rightleftharpoons \mathop {\mathrm{Cr^{3+}_{(aq)}}}\limits_s + \mathop {\mathrm{3OH^-_{(aq)}}}\limits_{3s} $

At equilibrium, the solubility product $\mathrm{K}_{\mathrm{sp}}$ is given by:

$ \mathrm{K}_{\mathrm{sp}} = (\mathrm{s}) \cdot (3 \mathrm{~s})^3 = 27 \mathrm{~s}^4 $

Substituting the given $\mathrm{K}_{\mathrm{sp}}$ value:

$ 27 \mathrm{~s}^4 = 1.6 \times 10^{-30} $

Solving for $\mathrm{s}$, we divide both sides by 27:

$ \mathrm{s} = \left(\frac{1.6 \times 10^{-30}}{27}\right)^{1/4} $

Thus, the molar solubility of $\mathrm{Cr}(\mathrm{OH})_3$ is:

$ \left(\frac{1.6}{27} \times 10^{-30}\right)^{1/4} $

When water is heated, its self-ionization increases. At 25°C, the water ionization constant is given by

$K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14},$

so in pure water,

$[\mathrm{H}^+] = [\mathrm{OH}^-] = 1.0 \times 10^{-7} \, \text{M},$

which corresponds to a pH of 7.

As water is heated to 80°C, the value of $K_w$ increases due to the endothermic nature of water’s autoionization. This means that both $[\mathrm{H}^+]$ and $[\mathrm{OH}^-]$ increase. However, since their concentrations remain equal (which defines neutrality at that temperature), the pH isn’t 7 anymore. Instead, the increased concentration of $\mathrm{H}^+$ ions leads to a pH value lower than 7.

Key points:

Heating water increases $K_w$.

Higher $K_w$ means higher $[\mathrm{H}^+]$ (and equally higher $[\mathrm{OH}^-]$).

Thus, the pH decreases (e.g., it might be around 6.5–6.6 at 80°C) even though the water is still neutral.

So, the correct answer is:

Option A – Decrease.

Condition for precipitation $\mathrm{Q}_{\mathrm{ip}}>\mathrm{K}_{\text {sp }}$

For $\left[\mathrm{A}(\mathrm{OH})_2\right.$]

$\begin{aligned} & {\left[\mathrm{A}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2>9 \times 10^{-10}} \\ & {\left[\mathrm{~A}^{+2}\right]=1 \mathrm{M}} \\ & \Rightarrow\left[\mathrm{OH}^{-}\right]>3 \times 10^{-5} \mathrm{M} \end{aligned}$

For $\left[\mathrm{B}(\mathrm{OH})_3\right]$

$\begin{aligned} & {\left[\mathrm{B}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3>27 \times 10^{-18}} \\ & {\left[\mathrm{~B}^{3+}\right]=1 \mathrm{M}} \\ & \Rightarrow\left[\mathrm{OH}^{-}\right]>3 \times 10^{-6} \mathrm{M} \end{aligned}$

So, $\mathrm{B}(\mathrm{OH})_3$ will precipitate before $\mathrm{A}(\mathrm{OH})_2$

$\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~s})^3 \cdot(4 \mathrm{~s})^4 \\\\ & \mathrm{~K}_{\mathrm{sp}}=6912 \mathrm{~s}^7 \\\\ & \mathrm{~s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}}\end{aligned}$

For a sparingly soluble salt $\mathrm{AB}_2$, the dissolution in water can be represented by the following equilibrium equation:

$ \mathrm{AB}_2 \rightleftharpoons \mathrm{A}^{2+} + 2\mathrm{B}^{-} $

When $\mathrm{AB}_2$ dissolves in water, it generates one $\mathrm{A}^{2+}$ ion and two $\mathrm{B}^{-}$ ions. The solubility product constant, $K_{sp}$, for this reaction can be expressed as:

$ K_{sp} = [\mathrm{A}^{2+}][\mathrm{B}^{-}]^2 $

Given in the problem, the equilibrium concentrations are:

$ [\mathrm{A}^{2+}] = 1.2 \times 10^{-4} \mathrm{M} $

$ [\mathrm{B}^{-}] = 0.24 \times 10^{-3} \mathrm{M} = 2.4 \times 10^{-4} \mathrm{M} $

Substituting these values into the $K_{sp}$ expression:

$ K_{sp} = (1.2 \times 10^{-4}) \times (2.4 \times 10^{-4})^2 $

Calculating $K_{sp}$:

$ K_{sp} = 1.2 \times 10^{-4} \times (5.76 \times 10^{-8}) $

$ K_{sp} = 6.912 \times 10^{-12} $

Therefore, the correct option is:

Option C: $6.91 \times 10^{-12}$

Let's analyze both statements given to determine the correct answer.

Statement (I) describes a buffer solution as a mixture of a salt and an acid or a base mixed in any particular quantities. However, this definition is partially incorrect. A buffer solution is more accurately defined as a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. The purpose of a buffer solution is to maintain a stable pH when small amounts of acid (H⁺ ions) or alkali (OH⁻ ions) are added. It is not just any mixture of a salt and an acid or base but must involve components that can react with added acid or base to minimize changes in pH.

Therefore, Statement (I) is misleading or incomplete as it omits the necessity for a weak acid and its conjugate base or a weak base and its conjugate acid to form a true buffer solution.

Statement (II) mentions that blood is a naturally occurring buffer solution whose pH is maintained by the $\mathrm{H}_2\mathrm{CO}_3/\mathrm{HCO}_3^{-}$ (carbonic acid/bicarbonate) system. This is accurate. The carbonic acid-bicarbonate buffering system is one of the main buffering systems in human blood, helping to maintain the pH within a narrow range (typically around 7.35 to 7.45). This buffer system works by the reversible reaction between carbon dioxide (CO₂) and water to form carbonic acid ($\mathrm{H}_2\mathrm{CO}_3$), which can then dissociate into bicarbonate ion ($\mathrm{HCO}_3^{-}$) and a hydrogen ion (H⁺). This system effectively moderates changes in pH by either consuming or releasing H⁺ ions.

Given this analysis:

Statement I is false because it inaccurately or incompletely describes a buffer solution.

Statement II is true as it correctly identifies blood as a naturally occurring buffer solution that uses the $\mathrm{H}_2\mathrm{CO}_3/\mathrm{HCO}_3^{-}$ system to maintain pH.

Therefore, the correct answer is:

Option C: Statement I is false but Statement II is true.

The equilibrium $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightleftharpoons 2 \mathrm{CrO}_4^{2-}$ can be influenced by changes in the pH of the medium, according to Le Chatelier's principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

In this case:

$\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is dichromate, which exists more prominently in acidic conditions. On the other hand, $\mathrm{CrO}_4^{2-}$ is chromate, which is favored in basic conditions. When the medium is basic, hydrogen ion concentration decreases, leading to the formation of more chromate ions ($\mathrm{CrO}_4^{2-}$).

Thus, in a basic medium, the reaction shifts to the right, resulting in the formation of $2 \mathrm{CrO}_4^{2-}$ from $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$.

Therefore, the correct option is:

Option B: a basic medium

To determine the solubility product (K_sp) of calcium phosphate, we need to consider its chemical formula and how it dissociates in water. The formula for calcium phosphate is $\text{Ca}_3(\text{PO}_4)_2$. When dissolved in water, it dissociates according to the following equation:

$ \text{Ca}_3(\text{PO}_4)_2 (s) \rightleftharpoons 3 \text{Ca}^{2+} (aq) + 2 \text{PO}_4^{3-} (aq) $

Let $S$ be the molar solubility of calcium phosphate in $\text{mol}/L$. According to the stoichiometry of the dissociation, if $S$ moles per liter of calcium phosphate dissolve, $3S$ moles per liter of calcium ions and $2S$ moles per liter of phosphate ions are produced.

The expression for the solubility product constant (Ksp) for calcium phosphate in water is the product of the concentrations of the ions raised to the power of their respective coefficients in the balanced equation:

$ K_{sp} = [\text{Ca}^{2+}]^3 [\text{PO}_4^{3-}]^2 $

Substituting the concentrations with $3S$ for $\text{Ca}^{2+}$ and $2S$ for $\text{PO}_4^{3-}$, we get:

$ K_{sp} = (3S)^3 (2S)^2 = 27S^3 \cdot 4S^2 = 108S^5 $

If the solubility of calcium phosphate is $\text{W}_{g}$ per $100 \text{ mL}$ of water at $25^\circ\text{C}$, we need to convert grams to moles to find $S$. Solubility in moles per liter (which is the same as moles per $1000 \text{ mL}$) is:

$ S = \frac{\text{W}}{\text{M}} \times \frac{1000 \text{ mL}}{100 \text{ mL}} = 10\frac{\text{W}}{\text{M}} $

Now, substituting $S$ with $10\frac{\text{W}}{\text{M}}$ into the expression for $K_{sp}$:

$ K_{sp} = 108 \left(10\frac{\text{W}}{\text{M}}\right)^5 $

This simplifies to:

$ K_{sp} = 108 \times 10^5 \left(\frac{\text{W}}{\text{M}}\right)^5 $

Since $108$ is on the order of $10^2$, we can approximate this to:

$ K_{sp} \approx 10^7 \left(\frac{\text{W}}{\text{M}}\right)^5 $

Therefore, the correct answer based on the options provided is:

Option C: $10^7 \left(\frac{\text{W}}{\text{M}}\right)^5$

The correct answer is Option A: Both Statement I and Statement II are correct.

Explanation:

Statement I explains that an aqueous solution of ammonium carbonate is basic. This is because ammonium carbonate ($\text{(NH}_4)_2\text{CO}_3$) dissociates in water to form ammonium ions ($\text{NH}_4^+$) and carbonate ions ($\text{CO}_3^{2-}$). The ammonium ion ($\text{NH}_4^+$) is a weak acid and can donate a proton to form ammonia ($\text{NH}_3$) and a hydronium ion ($\text{H}_3\text{O}^+$), but this happens to a minimal extent due to its weak acidic nature. On the other hand, the carbonate ion ($\text{CO}_3^{2-}$) can accept a hydrogen ion ($\text{H}^+$) from water to form bicarbonate ($\text{HCO}_3^-$) and hydroxide ions (${OH}^-$), which makes the solution basic. The reaction of carbonate ions removing hydrogen ions from water is more pronounced than the reaction of ammonium ions donating hydrogen ions to water, hence the overall solution becomes basic.

Statement II is about the general principle that determines the acidic or basic nature of a salt solution formed by the salt of a weak acid and a weak base. The acidic or basic nature of such a salt solution depends on the acid dissociation constant ($K_a$) of the acid and the base dissociation constant ($K_b$) of the base. If $K_a > K_b$, the solution tends to be acidic because the weak acid is stronger (more willing to donate protons) than the weak base is at accepting protons. Conversely, if $K_b > K_a$, the solution tends to be basic because the weak base is stronger (more willing to accept protons) than the weak acid is at donating protons. In the case of ammonium carbonate, carbonic acid ($H_2CO_3$) is a weak acid, and ammonia ($NH_3$) is a weak base. The basic nature of ammonium carbonate solution suggests that the $K_b$ of ammonia exceeds the $K_a$ of carbonic acid for the reactions in an aqueous solution, making the solution basic.

Therefore, both statements I and II are correct, thereby making Option A the right choice.

(A) The $\mathrm{pH}$ of $1 \times 10^{-8}~ \mathrm{M} ~\mathrm{HCl}$ solution is 8.

This statement is incorrect. For a strong acid like HCl, the concentration of H+ ions will be the same as the concentration of the acid, i.e., $1 \times 10^{-8}~\mathrm{M}$. The pH can be calculated using the formula:

$\mathrm{pH} = -\log [\mathrm{H}^+] = -\log (1 \times 10^{-8}) = 8$

However, because the concentration is so low, it approaches the range where water auto-ionization becomes significant. In this case, the solution pH will be slightly higher than 7, but not exactly 8.

(B) The conjugate base of $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ is $\mathrm{HPO}_{4}^{2-}$.

This statement is correct. The conjugate base of an acid is formed when it loses one H+ ion:

$\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}^{+}$

(C) $\mathrm{K}_{\mathrm{w}}$ increases with an increase in temperature.

This statement is correct. The ion product of water, $\mathrm{K}_{\mathrm{w}}$, increases with increasing temperature. This is because the auto-ionization of water is an endothermic process, meaning it absorbs heat:

$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{-}$

As the temperature increases, the equilibrium shifts towards the formation of more $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ ions, leading to an increase in $\mathrm{K}_{\mathrm{w}}$.

(D) When a solution of a weak monoprotic acid is titrated against a strong base at the half-neutralization point, $\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}$

This statement is incorrect. At the half-neutralization point, the concentration of the weak acid ([HA]) is equal to the concentration of its conjugate base ([A-]). According to the Henderson-Hasselbalch equation:

$\mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}$

At the half-neutralization point, the ratio of [A-] to [HA] is 1, so the equation becomes:

$\mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log (1) = \mathrm{pK}_{\mathrm{a}}$

Therefore, the correct answer is:

(B) and (C) are correct.

The reaction between silver nitrate (AgNO₃) and potassium iodide (KI) forms silver iodide (AgI), which is practically insoluble in water. The reaction is as follows :

AgNO₃(aq) + KI(aq) → AgI(s) + KNO₃(aq)

Although the reaction stoichiometry indicates that iodide ions (I^-) and silver ions (Ag⁺) are produced, silver iodide (AgI) precipitates out of the solution due to its low solubility, and very little Ag⁺ and I^- ions remain in the solution. Hence, their concentration in the solution will be negligible.

So, the correct answer should be :

Option C : Ag+ and I- both

Indicator	pH range
Methyl orange	3.2 - 4.5
Phenolpthalein	8.3 - 10.5

Option B is the incorrect statement for the use of indicators in acid-base titration.

Methyl orange is a suitable indicator for a strong acid vs strong base titration, not for a weak acid vs weak base titration. In a weak acid vs weak base titration, the pH at the equivalence point is typically around 7, which is within the pH range where methyl orange undergoes a color change. Therefore, methyl orange is not suitable for use in weak acid vs weak base titrations.

The other options are correct:

Phenolphthalein is a suitable indicator for a weak acid vs strong base titration because the pH at the equivalence point is basic, which causes phenolphthalein to undergo a color change.

Methyl orange is a suitable indicator for a strong acid vs weak base titration because the pH at the equivalence point is acidic, which causes methyl orange to undergo a color change.

Phenolphthalein may be used for a strong acid vs strong base titration because the pH at the equivalence point is basic, which causes phenolphthalein to undergo a color change.

Let the initial concentration of $\mathrm{H}^{+}$ be 1

$\therefore \left[\mathrm{H}^{+}\right]_{\mathrm{i}}=1 \Rightarrow \mathrm{pH}=0$

It changes by 1000 units

$\therefore \left[\mathrm{H}^{+}\right]_{\mathrm{f}}=10^{3} \Rightarrow \mathrm{pH}=-3$

$\therefore \mathrm{pH}$ decreases by 3 units

$\begin{aligned} {\left[\mathrm{H}^{+}\right] } &=\frac{0.01 \times 200+2 \times 0.01 \times 400}{600} \\ &=\frac{0.01+2 \times 0.01 \times 2}{3} \\ &=\frac{0.01+0.04}{3} \\ &=\frac{5}{3} \times 10^{-2} \\ \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\ &=-\log \left(\frac{5}{3} \times 10^{-2}\right) \\ &=-\left[\log \frac{5}{3}+\log 10^{-2}\right] \\ &=-[\log 5-\log 3-2] \\ &=-0.7+0.48+2 \\ &=2.48-0.7 \\ &=1.78 \end{aligned}$

$2 \mathrm{KMnO}_4+16 \mathrm{HCl} \rightarrow 2 \mathrm{MnCl}_2+2 \mathrm{KCl}+8 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2$

$\mathrm{HCl}$ is not used in the process of titration because it reacts with the $\left(\mathrm{KMnO}_{4}\right)$ that is used in the process and gets oxidized.

$\mathrm{NH}_{4} \mathrm{OH}$ is a weak base and $\mathrm{HCl}$ is a strong acid.

With the addition of $\mathrm{HCl}$ to $\mathrm{NH}_{4} \mathrm{OH}, \mathrm{pH}$ of solution will decrease gradually.

So, the correct graph should be

For basic Buffer, $\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { salt }]}{[\text { Base }]}$

$\mathrm{pOH}=14-8.26=5.74$

$5.74=4.74+\log \frac{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}{0.2}$

$\left[\mathrm{NH}_{4} \mathrm{Cl}\right]=2 \mathrm{M}$

Moles of $\mathrm{NH}_{4} \mathrm{Cl}=2 \times 1=2$ moles

Weight of $\mathrm{NH}_{4} \mathrm{Cl}=2 \times 53.5=107 \mathrm{~g}$

$\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \quad \mathrm{K}_{\mathrm{a}_{3}}$

$\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \quad \mathrm{K}_{\mathrm{a}_{1}}$

$\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \quad \mathrm{K}_{\mathrm{a}_{2}}$

$\mathrm{K}_{\mathrm{a}_{3}}=\frac{\left[\mathrm{H}^{+}\right]^{2}\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right]}$

$\mathrm{K}_{\mathrm{a}_{1}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right]}, \mathrm{K}_{\mathrm{a}_{2}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{-}\right]}{\left[\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\right]}$

$\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}} \times \mathrm{K}_{\mathrm{a}_{2}}$

$\left[\mathrm{NH}_4^{+}\right]=\frac{2 \mathrm{mmole}}{60 \mathrm{ml}}=\frac{1}{30} \mathrm{M}$

$\mathrm{pH}=\frac{\mathrm{pK}_{\mathrm{w}}-\mathrm{pK}_{\mathrm{b}}-\log \mathrm{C}}{2}=\frac{14-5+1.48}{2}=5.24$

The solubility of AgCl will be maximum in deionized water.

Silver chloride (AgCl) is a sparingly soluble salt. Its solubility in water can be represented by the equilibrium:

$ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) $

In the presence of a common ion, the solubility of AgCl decreases due to the common ion effect, which is explained by Le Chatelier's Principle. Here's an analysis of each option:

Option A: 0.01 M KCl

The presence of additional chloride ions ($\text{Cl}^-$) from KCl will shift the equilibrium to the left, reducing the solubility of AgCl.

Option B: 0.01 M HCl

Similar to KCl, HCl also provides more chloride ions, which will decrease the solubility of AgCl due to the common ion effect.

Option C: 0.01 M AgNO$_3$

AgNO$_3$ introduces additional silver ions ($\text{Ag}^+$) into the solution, also shifting the equilibrium to the left, hence decreasing the solubility of AgCl.

Option D: Deionized water

This option does not introduce additional ions that impact the equilibrium, so there are no additional silver or chloride ions to shift the equilibrium. As a result, the solubility of AgCl will be highest in deionized water, as there is no common ion effect.

Therefore, the solubility of AgCl is highest in deionized water.

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}+\mathrm{H}^{+}$

From Henderson equation

$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$

$4=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$

$-\log 10^{-4}=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}$

$-\log 10^{-4}=-\log 1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}$

$10^{-4}=1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}$

$\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}=0.13$

K_sp = (2s)²(3s)³ = 108s⁵

108s⁵ = 108 × 10^–75

s = 1.0 × 10^–15 mol/L

The amphoteric nature of water is explained by using Bronsted-Lowry acid base concept

$\underset{\text { (acid) }}{\mathrm{H}_{2} \mathrm{O}}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{OH}^{-}+\mathrm{NH}_{4}^{+}$

$\underset{\text { (base) }}{\mathrm{H}_{2} \mathrm{O}}+\mathrm{H}_{2} \mathrm{~S} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HS}^{-}$

Hence, A is false but R is true

Titration curve for strong acid and weak base initially a buffer of weak base and conjugate acid is :



Formed, thus pH falls slowly and after equivalence point, so the pH falls sharply so methyl orange, having pH range of 3.2 to 4.4 will weak as indicator. So, statement I is correct.



Titration curve for weak acid and strong base (NaOH)

Initially weak acid will form a buffer so pH increases slowly but after equivalence point. It rises sharply covering range of phenolphthalein so it will be suitable indicator so statement II is false.

Conc. of Cl^$-$ = 0.1 M = 10^$-$1 M

Conc. of CrO$_4^{2 - }$ = 0.001 M = 10^$-$3 M

K_sp(AgCl) = [Ag⁺][Cl^$-$]

[Ag⁺]_AgCl = ${{1.7 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}} = 1.7 \times {10^{ - 9}}$

${K_{sp}}(A{g_2}Cr{O_4}) = {[A{g^ + }]^2}[CrO_4^{2 - }]$

$[A{g^ + }] = \sqrt {{{1.9 \times {{10}^{ - 12}}} \over {{{10}^{ - 3}}}}} = \sqrt {19} \times {10^{ - 4}}$

$\therefore$ AgCl will be precipitate first

A : For temporary hardness,
Mg(HCO₃)₂ $\buildrel {heating} \over \longrightarrow $ Mg(OH)₂ + 2CO₂ Assertion is false. MgCO₃ has high solubility product than Mg(OH)₂.

R : MgCO₃ is more water soluble than Mg(OH)₂.

K_SP(Mg(OH)2) = 1.8 × 10^–11

K_SP(MgCO3) = 3.5 × 10^–8

NF₃ has no vacant orbital neither in nitrogen nor in fluorine so it cannot accept the electron & hence cannot acts as lewis acid and but for PCl₅ P has no L.P & hence it cannot acts as base but ClF3 (3 B.P + 2 L.P) & SF₄ (4 B.P + 1 L.P)

Let, solubility of Ca(OH)₂ in pure water = S mol/L

$Ca{(OH)_2}$ $\rightleftharpoons$ $\mathop {C{a^{2 + }}}\limits_{S\,mol/L} + \mathop {2O{H^ - }}\limits_{2 \times S\,(mol/L)} $

K_sp = [Ca²⁺] [OH^$-$]² = S $\times$ (2S)² = 4 S³ (mol/L)

The expression of K_sp can also be written as,

K_sp = x^x . y^y . S^{x + y}

= 1¹ . 2² . S^{1 + 2}

= 4 S³ [$\because$ For Ca(OH)₂ : x = 1, y = 2]

x and y are the coefficients of cations and anions respectively

$S = {\left( {{{{K_{sp}}} \over 4}} \right)^{1/3}} = {\left( {{{5.5 \times {{10}^{ - 6}}} \over 4}} \right)^{1/3}}$

$ = 1.11 \times {10^{ - 2}}$ ml/L

Let solubility is x



${K_{sp}} \times {1 \over {{K_a}}} = [A{g^ + }][C{N^ - }] \times {{[HCN]} \over {[{H^ + }][C{N^ - }]}}$

$2.2 \times {10^{ - 16}} \times {1 \over {6.2 \times {{10}^{ - 10}}}} = {{[S][S]} \over {{{10}^{ - 3}}}}$

${S^2} = {{2.2} \over {6.2}} \times {10^{ - 9}}$

${S^2} = 3.55 \times {10^{ - 10}}$

$S = \sqrt {3.55 \times {{10}^{ - 10}}} $

$S = 1.88 \times {10^{ - 5}} = 1.9 \times {10^{ - 5}}$

(A) 10^–2 M HCl $ \Rightarrow $ [H⁺] = 10^–2 M $ \Rightarrow $ pH = 2 $ \Rightarrow $ pOH = 14 - 2 = 12
(B) 10^–2 M NaOH $ \Rightarrow $ [OH^–] = 10^–2 M $ \Rightarrow $ pOH = 2
(C) 10^–2 M CH3COO^–Na⁺ $ \Rightarrow $ [OH⁺] > 10^–7 $ \Rightarrow $ pOH < 7
(D) 10^–2 M NaCl $ \Rightarrow $ Neutral pOH = 7

Order of pOH value A > D > C > B

Steep rise in pH around the equivalence point for titration of strong acid with strong base.

So finally we get mixture of

CH₃COOH + CH₃COONa that will work like acidic buffer solution.

K_sp value of CuS is very low 10^–36 (3.6 × 10^–36) due to low K_sp value Cu⁺² ion gets precipitated very quickly even with very low concentration of S^–2 ion.

Cr(OH)3	⇌	Cr+3	+	3OH-
		S		3S

K_sp = [Cr³⁺] [OH^– ]³

$ \Rightarrow $ 6 × 10^–31 = S × (3S)³

$ \Rightarrow $ 6 × 10^–31 = 27 S⁴

S = ${\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}$

As [OH^–] = 3S

= $3{\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}$

= (18 × 10^–31)^1/4 M

[Pb²⁺] = ${{300 \times 0.134} \over {400}}$

= 1.005 × 10^–1 M

[Cl^-] = ${{100 \times 0.4} \over {400}}$

= 10^–1 M

$PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - $

Q = [Pb²⁺] × [Cl^–]²

= 0.1005 × (0.1)²

= 1.005 × 10^–3

Given K_sp = 1.6 × 10^–5

$ \therefore $ Q > K_sp

H₂O ⇌ H⁺ + OH^-

As $\Delta $H $>$ 0 for this reaction. So Dissociation of H₂O is endothermic.

pH = $ - \ln \left[ {{H^ + }} \right]$

k_w = $\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$

$\ln \left\{ {{{{k_{{T_2}}}} \over {{k_{{T_1}}}}}} \right\} = {{\Delta H^\circ } \over R}\left\{ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right\}$

When temperature increases then

${{\Delta H^\circ } \over R}\left\{ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right\} > 0$, as reaction is endothermic.

$ \therefore $ k_T2 $>$ k_T1

$ \Rightarrow $ ${\left[ {{H^ + }} \right]_{{T_2}}}$ $>$ ${\left[ {{H^ + }} \right]_{{T_1}}}$

$ \therefore $ [pH]_T2 $<$ [pH]_T1

Aq. NaOH in a burette and aqueous oxalic acid in a conical flask.

From the given curve,

if [X] = 1 mM then [Y] = 2 mM

$ \therefore $ Salt is XY₂

XY₂(s) ⇌ X²⁺(aq.) + 2Y^-(aq.)

k_sp = [X²⁺][Y^–]²

= (10^–3) (2 × 10^–3)²

= 4 × 10^–9 M³

0.1 M Formic acid(HCOOH) (A),
0.1 M Acetic acid(CH₃COOH) (B),
0.1 M Benzoic acid(C₆H₅COOH) (C)

Conductivity (K) depends on no of ions present in unit volume of solution. When no of ions increases conductivity also increases.
Also no of ions depends on degree of dissociation($\alpha $).
We know, $\alpha $ = $\sqrt {{{{K_a}} \over C}} $

For all solutions C = 0.1 M
So, $\alpha $ depends on only K_$a$ here.
K_$a$ of HCOOH = 10^-4
K_$a$ of CH₃COOH = 1.8 $ \times $ 10^-5
and K_$a$ of C₆H₅COOH = 6 $ \times $ 10^-5

$ \therefore $ $\alpha $(HCOOH) > $\alpha $(C₆H₅COOH) > $\alpha $(CH₃COOH)

Therefore conductivity order = A > C > B

Cd(OH)2(s)	⇌	Cd2+(aq)	+	2OH$-$(aq)
		S		2S

$ \therefore $ K_sp[Cd(OH)₂] = 4(S)³ = 4(1.84 × 10^–5)³ [Given S = 1.84 × 10^–5]

For buffer solution of pH = 12 :
As pH = 12 so pOH = 2
$ \Rightarrow $ [OH^$-$] = 10^-2
Let the solubilty = S₁

Cd(OH)2(s)	⇌	Cd2+(aq)	+	2OH$-$(aq)
		S1		2S1 + 10-2

$ \therefore $ K_sp[Cd(OH)₂] = S₁(10^-2)² = 4(1.84 × 10^–5)³
$ \Rightarrow $ S₁ = 2.49 × 10^–10 M

${K_{sp}} = $ [Al³⁺][OH^-]³

$ \Rightarrow $ 2.4 × 10^–24 = s(0.2)³

$ \Rightarrow $ s = ${{2.4 \times {{10}^{ - 24}}} \over {8 \times {{10}^{ - 3}}}}$

$ \Rightarrow $ s = 3 × 10^–22

NH₄⁺ + H₂O ⇋ NH₄OH + H⁺

[H⁺] = c$\alpha $

= $\sqrt {{k_a}\left( {NH_4^ + } \right) \times c} $

= $\sqrt {{{{k_w}} \over {{k_b}\left( {N{H_4}OH} \right)}} \times c} $

= $\sqrt {{{{{10}^{ - 14}}} \over {{{10}^{ - 5}}}} \times 0.02} $

= $\sqrt {20} \times {10^{ - 6}}$

$ \therefore $ pH = $ - \log \left( {\sqrt {20} \times {{10}^{ - 6}}} \right)$ = 5.35

(a) Equivalance of strong acid = 0.1 $ \times $ 2 $ \times $ 400 = 80

Equivalance of strong base = 0.1 $ \times $ 400 = 40

$ \therefore $ [H⁺] of mixture = ${{80 - 40} \over {800}}$ = ${1 \over {20}}$

$ \therefore $ pH = $ - \log \left[ {{H^ + }} \right]$ = $ - \log \left( {{1 \over {20}}} \right)$ = 1.3

(b) Ionic product of water increases with increase of temperature because ionisation of water is endothermic.

(c)

k_a = ${{{{10}^{ - 5}} \times c\alpha } \over {c\left( {1 - \alpha } \right)}}$

$ \Rightarrow $ 10^-5 = ${{{{10}^{ - 5}} \times \alpha } \over {\left( {1 - \alpha } \right)}}$

$ \Rightarrow $ ${\alpha \over {\left( {1 - \alpha } \right)}}$ = 1

$ \Rightarrow $ $\alpha $ = 0.5

$ \therefore $ Degree of dissociation($\alpha $) = 50%

(d) The Le Chatelier's principle is always applicable to common-ion effect.