Question

The $pH$ of $0.01\, M\, HCOOH$ will be_____.
[Given: $\sqrt{ K _{ a }}=1.18 \times 10^{-2}, \left.\log _{10} 1.18=0.0719\right]$

Answer 1

For weak acid,
$\left[ H ^{+}\right]=\sqrt{ K _{ a } \times C }$
$=\sqrt{ K _{ a }} \times \sqrt{ C }$
$=1.18 \times 10^{-2} \times \sqrt{0.01}$
$=1.18 \times 10^{-2} \times \sqrt{10^{-2}}$
$=1.18 \times 10^{-2} \times 10^{-1}$
$=1.18 \times 10^{-3}$
$pH =\log _{10}\left[ H ^{+}\right]$
$=-\log _{10}\left[1.18 \times 10^{-3}\right]$
$=-\left[\log _{10} 1.18+\log _{10}\left(10^{-3}\right)\right]$
$=-[0.0719-3]=-[-2.9281]=2.93$