Q38 / 160 · AP EAMCET 2009 · Chemistry

$20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ acetic acid is mixed with $50 \mathrm{~mL}$ of potassium acetate. $K_a$ of acetic acid $=1.8 \times 10^{-5}$ at $27^{\circ} \mathrm{C}$. Calculate concentration of potassium acetate if $\mathrm{pH}$ of the mixture is 4.8 .

  • A
    0.1 M
  • Correct answer
    B
    0.04 M
  • C
    0.4 M
  • D
    0.02 M

Solution

Let the concentration of potassium acetate is $x$. From Henderson's equation, $\begin{aligned} & \mathrm{pH}=\mathrm{p} K_a+\log \frac{[\text { salt }]}{[\text { acid }]} \\ & 4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{x \times 50}{20 \times 0.1 \mathrm{M}} \\ & 4.8=4.74+\log 25 x \end{aligned}$ $\begin{array}{rlrl} \text { or } & \log 25 x=0.06 \\ \therefore 25 x & =1.148 \\ & x=0.045 \mathrm{M} \end{array}$

Hence, the correct answer is option: 'B'.

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