Question

Find the pH value of 0.01 M $C{H}_{3}COOH$ at ${25}^o$ C.

• A. 2
• B. 12
• C. 1
• D. None of the above

Answer 1

The correct option is A 2

Given concentration of weak acid ${CH}_{3}COOH$ : $C_{weak}=$$0.01M$

We know the $K_a$ of ${CH}_{3}COOH$ i.e, ${1.86\times 10}^{-5}$

Concentration of Hydrogen ion $\left[H^{+}\right]$ $=$ $K_a\frac{C_{weak}-\left[H^{+}\right]}{\left[H^{+}\right]}$

${\left[H^{+}\right]}^2=K_a(C_{weak}-\left[H^{+}\right])$

${\left[H^{+}\right]}^2+K_a\left[H^{+}\right]-K_a{C}_{weak}=0$

By solving this equation we get, $\left[H^{+}\right]=\frac{{-K}_a\pm \sqrt{{K_a}^2+4\left(K_a{C}_{weak}\right)}}{2}$

Substituting the values of $K_a$ and $C_{weak}$ we get,

$\left[H^{+}\right]=\frac{{-1.86\times 10}^{-5}\pm \sqrt{{{(1.86\times 10}^{-5})}^2+4({1.86\times 10}^{-5}\times 0.01)}}{2}$

$\left[H^{+}\right]={4.22077\times 10}^{-4}M$ (we take only positive value as concentrtion cannot have negative value)

Now we know that $pH=-\log \left[H^{+}\right]=-\log ({4.22077\times 10}^{-4})=3.375$