Q45 / 170 · Manipal (MET) 2018 · Chemistry

Degree of dissociation of an acid HCl is $95 \%$. 0.192 g of the acid is present in 0.5 L of solution. The pH of the solution is

  • Correct answer
    A
    2
  • B
    1
  • C
    3
  • D
    4

Solution

Number of moles of \(\mathrm{HCl}=\frac{0.192}{36.5}=0.0053\) Volume \(=0.5 \mathrm{~L}\) Molarity of \(\mathrm{HCl}=\frac{0.0053}{0.5}=1.06 \times 10^{-2} \mathrm{M}\) \(\left[\mathrm{H}^{+}\right]\)in HCl solution \(=1.06 \times 10^{-2} \times \frac{95}{100} \approx 1 \times 10^{-2}\) \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-2}\right)=2\)

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

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