Buffer solution

Chemistry Level 2

There is a 500 mL solution of 0.1 M acetic acid. How much of 0.05 M NaOH needs to be added to produce a buffer of pH 4.75, given that the p K a \text{p}K_a of acetic acid is 4.75?

250 mL 250\text{ mL} 500 mL 500\text{ mL} 750 mL 750\text{ mL} 1000 mL 1000\text{ mL}

Krishna Singh by Krishna Singh , 22, India

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1 solution

Sowmitra Das
May 12, 2014

From the formula of p H pH of a Buffer solution,
p H = p K a + log ⁡ [ S a l t ] [ A c i d ] pH=pK_a+\log{\frac{[Salt]}{[Acid]}}
⇒ 4.75 = 4.75 + log ⁡ [ S a l t ] [ A c i d ] \Rightarrow 4.75=4.75 + \log{\frac{[Salt]}{[Acid]}}
⇒ [ S a l t ] = [ A c i d ] \Rightarrow [Salt]=[Acid]
⇒ m ( S a l t ) = m ( A c i d ) \Rightarrow m(Salt)=m(Acid) ......(1)
where, m ( x ) m(x) means the no. of moles of x x .

N a O H NaOH reacts with C H 3 C O O H CH_3COOH as follows:
N a O H + C H 3 C O O H → C H 3 C O O N a + H 2 O NaOH+CH_3COOH \rightarrow CH_3COONa+H_2O

So, 1 1 mole N a O H ≡ NaOH\equiv 1 1 mole C H 3 C O O H ≡ 1 CH_3COOH\equiv 1 mole C H 3 C O O N a CH_3COONa

Suppose, V V litres of N a O H NaOH is required.
∴ \therefore No. of moles of C H 3 C O O N a CH_3COONa formed = 0.05 × V =0.05\times V
and, No. of moles of Acetic Acid left = 0.1 × 0.5 − 0.05 × V =0.1\times 0.5-0.05\times V

From (1),
0.05 × V = 0.1 × 0.5 − 0.05 × V 0.05\times V=0.1\times 0.5-0.05\times V
⇒ V = 0.5 L = 500 m L \Rightarrow V=0.5 L=\boxed{500mL}

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