JEE MAIN - Chemistry (2017 - 9th April Morning Slot - No. 22)
50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be :
3.75
4.75
8.25
9.25
Explanation
NH3
+ HCl $$ \to $$ NH4Cl
moles of HCl = 0.2 M × 25 × 10–3 L = 0.005 moles HCl (total consumed)
moles of NH3 = 0.2 M × 50 × 10–3 L = 0.01 moles HCl
excess NH3 = 0.01 – 0.005 = 0.005 moles
From reaction, 1 mole ammonia = 1 mole NH4Cl
$$ \therefore $$ 0.005 NH3 = 0.005 NH4Cl
Total Volume = VHCl + VNH3 = 25 + 50 = 75 mL
[NH3] = [NH4Cl] = $${{0.005} \over {75 \times {{10}^{ - 3}}}}$$ = 0.066 M
pOH = pKb + log $${{\left[ {salt} \right]} \over {\left[ {base} \right]}}$$
= pKb + log$${{\left[ {N{H_4}Cl} \right]} \over {\left[ {N{H_3}} \right]}}$$
= 4.75 + log$${{\left[ {0.066} \right]} \over {\left[ {0.066} \right]}}$$
$$ \Rightarrow $$ pOH = 4.75
$$ \therefore $$ pH = 14 – 4.75 = 9.25
moles of HCl = 0.2 M × 25 × 10–3 L = 0.005 moles HCl (total consumed)
moles of NH3 = 0.2 M × 50 × 10–3 L = 0.01 moles HCl
excess NH3 = 0.01 – 0.005 = 0.005 moles
From reaction, 1 mole ammonia = 1 mole NH4Cl
$$ \therefore $$ 0.005 NH3 = 0.005 NH4Cl
Total Volume = VHCl + VNH3 = 25 + 50 = 75 mL
[NH3] = [NH4Cl] = $${{0.005} \over {75 \times {{10}^{ - 3}}}}$$ = 0.066 M
pOH = pKb + log $${{\left[ {salt} \right]} \over {\left[ {base} \right]}}$$
= pKb + log$${{\left[ {N{H_4}Cl} \right]} \over {\left[ {N{H_3}} \right]}}$$
= 4.75 + log$${{\left[ {0.066} \right]} \over {\left[ {0.066} \right]}}$$
$$ \Rightarrow $$ pOH = 4.75
$$ \therefore $$ pH = 14 – 4.75 = 9.25
Comments (0)
