Browse Questions

# 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

(a) 2

(b) 1

(c) 3

(d) 4

Toolbox:
• $\rho H = - log [H^+]$

No. of moles of $HCl = \frac{0.192}{36.5}$ = 0.0053
Volume = 0.5 litre

Molarity of $HCL = \frac{0.0053}{0.5} = 1.06 \times 10^{-2}$ M

$[H^+]$ in $HCL$ solution = $1.06 \times 10^{-2} \times \frac{95}{100} = 100.7 \times 10^{-4} \approx 1 \times 10^{-2}$

Thus, $\rho H = - log [H^+] = - log (1 \times 10^{-2}) = 2$