Ionic Equilibrium

$\mathrm{HCl}<\mathrm{HNO}_2<\mathrm{CH}_3 \mathrm{COOH}$

HCl is a strong acid (complete ionisation), $\mathrm{HNO}_2$ is a stronger weak acid than $\mathrm{CH}_3 \mathrm{COOH}$.

So, HCl has lowest $\mathrm{pH}, \mathrm{CH}_3 \mathrm{COOH}$ highest pH .

Order of acidity: $\mathrm{HCl}>\mathrm{HNO}_2>\mathrm{CH}_3 \mathrm{COOH}$

Hence, the correct order of $\mathrm{pH}: \mathrm{HCl}<\mathrm{HNO}_2 <\mathrm{CH}_3 \mathrm{COOH}$

$ \begin{aligned} &\text { Henderson equation, } \mathrm{pH}=\mathrm{p} K_a+\log \left(\frac{[\text { Salt }]}{[\text { Acid }]}\right)\\ &\begin{aligned} \mathrm{pK}_a & =-\log \left(1.8 \times 10^{-5}\right)=4.74 \\ \text { Ratio } & =\frac{(0.1 \times 60)}{(0.1 \times 30)}=2 \\ \mathrm{pH} & =4.74+\log (2) \\ & \approx 4.74+0.3=5.04 \end{aligned} \end{aligned} $

Stronger the acid, lower is the pH or weaker the acid, higher the pH .

The strength of the given acid is in the order

$ \mathrm{HCl} > \mathrm{HCOOH} > \mathrm{CH}_{3} \mathrm{COOH} $

Hence, their pH is in the order

$ \mathrm{CH}_{3} \mathrm{COOH} > \mathrm{HCOOH} > \mathrm{HCl} $

For an acidic buffer, we use the Henderson-Hasselbalch equation:

$ \text{pH} = \text{p}K_a + \log \frac{[\text{Salt}]}{[\text{Acid}]} $

Let $ x $ be the concentration of the potassium acetate solution in molarity (M).

Calculate the moles of acetic acid:

$ 20 \, \text{mL} $ of $ 0.1 \, \text{M} $ acetic acid contains:

$ 20 \times 0.1 = 2 \, \text{millimoles} $

Calculate the moles of potassium acetate:

$ 50 \, \text{mL} $ of $ x \, \text{M} $ potassium acetate contains:

$ 50x \, \text{millimoles} $

Given the pH is 4.8, substitute the values into the Henderson-Hasselbalch equation:

$ 4.8 = -\left(\log 1.8 \times 10^{-5}\right) + \log \frac{50x}{2} $

Solve for $ x $:

$ \log 25x = 0.06 $

$ x = 0.04 \, \text{M} $

Thus, the concentration of potassium acetate is $ 0.04 \, \text{M} $.

The pH of a solution is calculated using the hydrogen ion concentration $[H^+]$ with the following formula:

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

Given the hydrogen ion concentration in the sample of soft drink is $3.8 \times 10^{-3} \mathrm{M}$.

To find the pH, we substitute the given $[H^+]$ value into the formula:

$\mathrm{pH} = -\log(3.8 \times 10^{-3})$

Breaking it down, we have:

$\mathrm{pH} = -\left(\log(3.8) + \log(10^{-3})\right)$

Recall that $\log(10^{-3}) = -3$, thus:

$\mathrm{pH} = -\left(\log(3.8) - 3\right)$

$\mathrm{pH} = -\log(3.8) + 3$

Now, finding the logarithm value of 3.8 using a calculator:

$\log(3.8) \approx 0.58$

Hence:

$\mathrm{pH} = -0.58 + 3$

$\mathrm{pH} \approx 2.42$

Thus, the pH of the soft drink is approximately 2.42.

Therefore, the correct option is: Option B (2.42)

$ \mathrm{NaOH} \rightleftharpoons \underset{(C)}{\mathrm{Na}^{+}}+\underset{(C)}{\mathrm{OH}^{-}} $

$ \underset{(x)}{\mathrm{Ni}(\mathrm{OH})_2} \rightleftharpoons \underset{(x)}{\mathrm{Ni}^{2+}}+\underset{(x)}{2 \mathrm{OH}^{-}}$

Total $\left[\mathrm{OH}^{-}\right]=x+C$

$\begin{aligned} \therefore \quad K_{\mathrm{sp}} & =\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & =x(x+C)^2=x\left(x^2+2 C x+C^2\right) \\ \end{aligned}$

or $K_{\text {sp }}=x C^2$ (neglecting higher powers of $x$)

$\begin{aligned} \Rightarrow \quad x & =\frac{K_{s p}}{C^2}=\frac{1.9 \times 10^{-15}}{(1)^2} \\ & =1.9 \times 10^{-15} \mathrm{M} \end{aligned}$

Sodium formate is present at the equivalence point. It is the salt of weak acid + strong base. So, final solution will be basic in nature.

As we know,

$pH = 7 + {{p{K_a}} \over 2} + {{\log C} \over 2}$

where, C is concentration of salt.

Total volume of solution = 100 + 100 = 200 mL

Concentration of salt (C) $ = 2 \times {{100} \over {200}} = 1$ M

$pH = 7 + {{3.74} \over 2} + {{\log [1]} \over 2}$

$pH = 7 + 1.87 + 0$

$pH = 8.87$

Pb(OH)₂ $\rightleftharpoons$ Pb² + 2OH^$-$

K_sp = [Pb²⁺] [OH^$-$]² = S $\times$ (2S)²

K_sp = 4S³ = 4 $\times$ (6.7 $\times$ 10^$-$6)³ = 1.20 $\times$ 10^$-$15

In a solution with pH = 8

[H⁺] = 10^$-$8 and [OH^$-$] = 10^$-$6

1.20 $\times$ ^$-$15 = [Pb²⁺] [10^$-$6]²

$[P{b^{2 + }}] = {{1.2 \times {{10}^{ - 15}}} \over {{{[{{10}^{ - 6}}]}^2}}} = 1.2 \times {10^{ - 3}}$ M