Ionic Equillibrium — JEE Main practice

1. For a basic buffer of $\mathrm{NH_4OH}$ and its salt $\mathrm{NH_4Cl}$:

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

2. Given $\mathrm{p}K_b=4.75$ and required pH = 9.25:

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

3. Substituting: $4.75=4.75+\log\left(\frac{[\text{salt}]}{[\text{base}]}\right) \;\Rightarrow\; \log\left(\frac{[\text{salt}]}{[\text{base}]}\right)=0 \;\Rightarrow\; \frac{[\text{salt}]}{[\text{base}]}=1$

4. So we need equal moles of salt and base after the reaction $\mathrm{NH_4OH+HCl\rightarrow NH_4Cl+H_2O}$. If $n_b$ (base) and $n_a$ (acid) are the initial moles, the condition for equal salt and remaining base is:

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

5. Checking Option B: $n_b=0.2\times0.5=0.10$ mol, $n_a=0.1\times0.5=0.05$ mol. Indeed $n_b=2n_a$ ($0.10=2\times0.05$).

6. After reaction: salt = 0.05 mol, base left = $0.10-0.05=0.05$ mol — equal, confirming this is the required buffer with $\mathrm{pOH}=\mathrm{p}K_b=4.75$ and $\mathrm{pH}=9.25$.

Hence, the answer is Option B.

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

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

2. Find $\mathrm{p}K_a$ of $BH^+$ using $\mathrm{p}K_w=14$ at 25°C:

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

3. Substituting into the buffer equation with pH = 9:

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

4. Since $\log5=0.699$:

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

5. Compute the moles: $n(\mathrm{HCl})=0.02\cdot\frac{x}{1000}$ and $n(B)_{\text{initial}}=0.02\cdot\frac{y}{1000}$. The reaction $B+\mathrm{HCl}\rightarrow BH^++Cl^-$ gives:

$$n(BH^+)=0.02\cdot\frac{x}{1000},\qquad n(B)=0.02\cdot\frac{(y-x)}{1000}$$

6. So:

$$\frac{[B]}{[BH^+]}=\frac{y-x}{x}=5 \;\Rightarrow\; y-x=5x \;\Rightarrow\; y=6x$$

7. Using the total volume condition $x+y=100$ and substituting $y=6x$:

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

8. So:

$$y=100-x=85.7143$$

Hence, $x=14.3$ mL and $y=85.7$ mL — Option B.

1. Let $s_1$ be the molar solubility of $\mathrm{Ag_2CrO_4}$. It dissociates as $\mathrm{Ag_2CrO_4\rightleftharpoons2Ag^++CrO_4^{2-}}$, giving $[\mathrm{Ag^+}]=2s_1$ and $[\mathrm{CrO_4^{2-}}]=s_1$.

2. So:

$$K_{sp}=[\mathrm{Ag^+}]^2[\mathrm{CrO_4^{2-}}]=(2s_1)^2(s_1)=4s_1^3$$

3. Given $K_{sp}=32x$:

$$32x=4s_1^3 \;\Rightarrow\; s_1^3=8x \;\Rightarrow\; s_1=2\sqrt[3]{x}$$

4. Let $s_2$ be the molar solubility of $\mathrm{AgBr}$, dissociating as $\mathrm{AgBr\rightleftharpoons Ag^++Br^-}$, giving $[\mathrm{Ag^+}]=[\mathrm{Br^-}]=s_2$.

5. So:

$$K_{sp}=[\mathrm{Ag^+}][\mathrm{Br^-}]=s_2^2$$

6. Given $K_{sp}=4y$:

$$4y=s_2^2 \;\Rightarrow\; s_2=2\sqrt{y}$$

7. The required ratio is:

$$\frac{s_1}{s_2}=\frac{2\sqrt[3]{x}}{2\sqrt{y}}=\frac{\sqrt[3]{x}}{\sqrt{y}}$$

Hence, the answer is Option B.