$20 m L$ of $0.2 M$ sodium hydroxide is added to $50 m L$ of $0.2 M$ acetic acid to give $70 m L$ of the solution. Calculate the $p H$ of the solution.
The dissociation constant for $C H_{3} C O O H$ is $1.8 \times 10^{- 5}$ .

3.56

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4.56

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5.56

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6.56

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Solution

The correct option is B 4.56
We know,
millimoles = molarity $\times$ volume (mL)
millimoles of $N a O H = 0.2 \times 20 = 4$
millimoles of $C H_{3} C O O H = 0.2 \times 50 = 10$

$C H_{3} C O O H (a q) + N a O H (a q) \to C H_{3} C O O N a (a q) + H_{2} O (l) I n i t i a l : 10400 E q u i l i b r i u m : (10 - 4) 044$

$[C H_{3} C O O N a] = \frac{4}{70} M [C H_{3} C O O H] = \frac{6}{70} M$
$p K_{a} = - l o g (K_{a}) p K_{a} = - l o g (1.8 \times 10^{- 5}) p K_{a} = (5 - 0.26) = 4.74$
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]} p H = p K_{a} + l o g (\frac{[C H_{3} C O O N a]}{[C H_{3} C O O H]}) p H = 4.74 + l o g ⎛ ⎝ \frac{\frac{4}{70}}{\frac{6}{70}} ⎞ ⎠ p H = 4.74 + l o g (0.67) p H = 4.74 - 0.18 = 4.56 p H = 4.56$

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Similar questions

20 m L

0.2 M

sodium hydroxide is added to

50 m L

0.2 M

acetic acid to give

70 m L

of the solution. What is the pH of the solution? Calculate the additional volume of

0.2 M

N a O H

required to make the pH of solution

4.74

. The ionisation constant of acetic acid is

1.8 \times 10^{- 5}

20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid to give 70 mL of the solution. What is the $p H$ of the solution? Calculate the additional volume $x$ (in mL) of 0.2 M NaOH required to make pH of the solution 4.74. The ionisation constant of acetic acid is $1.8 \times 10^{- 5}$ .