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Acidic Buffer Action

A 50 mL solut...

Question

A $50 m L$ solution of $0.2 M C H_{3} C O O H$ is titrated with $0.2 M N a O H$ . Find the $p H$ of solution when:
(a) $0 m L N a O H$ is added ( $p H_{i n i t i a l}$ )
(b) $10 m L$ of $N a O H$ has been added ( $p H_{10 m L}$ )

Given: $K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5}$

p H_{i n i t i a l} = 2.72, p H_{10 m L} = 4.14

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p H_{i n i t i a l} = 5.7, p H_{10 m L} = 6.72

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p H_{i n i t i a l} = 1.72, p H_{10 m L} = 7.28

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p H_{i n i t i a l} = 4.57, p H_{10 m L} = 2.17

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Solution

The correct option is A $p H_{i n i t i a l} = 2.72, p H_{10 m L} = 4.14$
Case(a)
mmol of $C H_{3} C O O H = 50 \times 0.2 = 10$
mmol of $N a O H = 0$
$C H_{3} C O O H (a q) + N a O H (a q) \to C H_{3} C O O N a (a q) + H_{2} O (l) 10000$
Since only weak acid $C H_{3} C O O H$ is present initially.
$p H_{i n i t i a l} = - l o g \sqrt{K_{a} C} p H_{i n i t i a l} = - \frac{1}{2} (l o g (1.8 \times 10^{- 5} \times 0.2)) p H_{i n i t i a l} = - \frac{1}{2} (l o g (3.6 \times 10^{- 6})) p H_{i n i t i a l} = \frac{1}{2} (6 - 0.56) = \frac{5.44}{2} p H_{i n i t i a l} = 2.72$

Case (b)
mmol of $C H_{3} C O O H = 50 \times 0.2 = 10$
mmol of $N a O H = 10 \times 0.2 = 2$

$C H_{3} C O O H (a q) + N a O H (a q) \to C H_{3} C O O N a (a q) + H_{2} O (l) I n i t i a l l y : 10200 F i n a l : 8022$

After titration we can see that $8 m m o l$ of $C H_{3} C O O H$ and $2 m m o l$ of $C H_{3} C O O N a$ are present.
$Final Concentration = \frac{Moles}{Total Volume}$
Total Volume $= 50 + 10 = 60 m L$
$[C H_{3} C O O H] = \frac{8}{60} M$
$[C H_{3} C O O N a] = \frac{2}{60} M$
An acidic buffer is formed. So,
$p H = p K_{a} + l o g (\frac{[C H_{3} C O O N a]}{[C H_{3} C O O H]}) p H = 4.74 + l o g ⎛ ⎝ \frac{\frac{2}{60}}{\frac{8}{60}} ⎞ ⎠ p H = 4.74 + l o g (\frac{1}{4}) = 4.74 - 0.6 p H_{10 m L} = 4.14$

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A 50 mL solution of 0.2 M CH3COOH is titrated with 0.2 M NaOH. Find the pH of solution when: (a) 0 mL NaOH is added (pHinitial) (b) 10 mL of NaOH has been added (pH10mL) Given: Ka(CH3COOH)=1.8×10−5

A $50 m L$ solution of $0.2 M C H_{3} C O O H$ is titrated with $0.2 M N a O H$ . Find the $p H$ of solution when:
(a) $0 m L N a O H$ is added ( $p H_{i n i t i a l}$ )
(b) $10 m L$ of $N a O H$ has been added ( $p H_{10 m L}$ )

Given: $K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5}$