Ionic equilibrium

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By Expert Faculty of Sri Chaitanya

Question

100 mL of a buffer solution contains 0.1 M each of weak acid HA and salt NaA. How many gram of NaOH should be added to the buffer so that it pH will be 6?
(K_aof HA = 10^-5)

Moderate

Solution

$For acidic buffer, pH = p K_{a} + \log \frac{0.1}{0.1}$
$pH = {pK}_{a} = - \log (10^{- 5}) = 5$
Rule: ABA (In acidic buffer (A), on addition of S_B (B), the concentration of W_A(A) decreases and that of salt increases.
Let x M of NaOH is added.
$\begin{array}{l} {pH}_{new} = 5 + \log (\frac{0 .1 + x}{0 .1 - x}) \\ 6 - 5 = \log (\frac{0 .1 + x}{0 .1 - x}) \\ (\frac{0 .1 + x}{0 .1 - x}) = Antilog (1) = 10 \end{array}$
Solve for x:
$\begin{array}{l} x = 0 .082 M = \frac{0 .082}{1000} \times 100 \\ = 0 .0082 mol (100 mL)^{- 1} \\ = 0 .0082 \times 40 g (100 mL)^{- 1} \\ = 0 .328 g \end{array}$

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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.