pH of a solution made by mixing $$50$$ mL of $$0.2$$ M $$NH4Cl$$ and $$75$$ mL of $$0.1$$ M NaOH is [pKb of $$NH3(aq)$$ $$=$$ $$4.74$$. log $$3$$ $$=$$ $$0.47$$]

$$N+H4$$ $$+$$ NaOH ⟶ $$NH4OH+NaCl$$ mmol $$50$$ × $$0.2$$ $$75$$ × $$0.1$$ $$=10$$ $$=75$$ mmol $$10-7.5$$ $$-$$ $$7.5$$ $$=2.5$$⇒This will result in a basic buffer.⇒$$pOH=pKb+log$$⁡[ Salt ][ Base ]$$=4.74+log$$⁡$$2.57.5=4.27$$⇒$$pH=14$$−$$4.27=9.73$$

Ionic equilibrium

Unlock the full solution & master the concept.

Get a detailed solution and exclusive access to our masterclass to ensure you never miss a concept

By Expert Faculty of Sri Chaitanya

Question

pH of a solution made by mixing 50 mL of 0.2 M NH₄Cl and 75 mL of 0.1 M NaOH is [pK_b of NH₃(aq) = 4.74. log 3 = 0.47]

Moderate

Solution

$\overset{+}{N} H_{4} + NaOH ⟶ {NH}_{4} OH + NaCl mmol 50 \times 0.2 75 \times 0.1 = 10 = 75 mmol 10 - 7.5 - 7.5 = 2.5$
$\Rightarrow$ This will result in a basic buffer.
$\begin{array}{l} \Rightarrow pOH = {pK}_{b} + \log \frac{[Salt]}{[Base]} \\ = 4 .74 + \log \frac{2 .5}{7 .5} = 4 .27 \\ \Rightarrow pH = 14 - 4 .27 = 9 .73 \end{array}$

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

Talk to our academic expert!

Create Your Own Test
Your Topic, Your Difficulty, Your Pace

Create Your Own Test
Your Topic, Your Difficulty, Your Pace

Similar Questions

Higher the amount of acid or base used to produce a definite change of pH in a buffer solution, higher will be its buffer capacity.Buffer capacity is maximum under the following conditions.

[Salt] = [Acid] for acid buffer solution and [Salt] = [Base] for basic buffer solution.p^H of a buffer lies in the range of $p^{k_{a}} \pm 1$ . Any buffer solution can be used as buffer up to two units depending on $p^{k_{a}}$ or $p^{k_{b}}$ values. When p^H = $p^{k_{a}}$ or $p^{k_{b}}$ solution is said to be efficient buffer.