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  4. 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

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Q. 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

Equilibrium

Solution:

No. of moles of $HCl=\frac{0.192}{36.5}=0.0053$
Volume $= 0.5$ It
Molarity of $HCl =\frac{0.0053}{0.5}=1.06\times10^{-2}\,M$
$[H^{+}]$ in $Hcl$ solution
$=1.06\times10^{-2}\times\frac{95}{100}$
$=100.7\times10^{-4}=1\times10^{-2}$
Thus $pH=-log[H]^{+}$
$=-log(1\times10^{-2})=2$