Question

Find out the $O{H}^{-}$ ion concentration in 100 mL of 0.015 M $HCl$ is:

• A. $2.0\times {10}^{-9}$ M
• B. $6.61\times {10}^{-13}$ M
• C. $3.0\times {10}^{-10}$ M
• D. $5\times {10}^{-12}$ M

Answer 1

The correct option is B $6.61\times {10}^{-13}$ M

$pH+pOH=14$

This is crucial since it connects $pH$ and $pOH$ of a given solution. Lets say that the $pH$ is $1$ then the $pOH$ will be $13$.

$pH=-log\left[H_3{O}^{+}\right]$

$pOH=-log\left[{OH}^{-}\right]$

$HCl$ is completely ionized in water because $HCl$ is a strong acid. So the $H_3{O}^{+}$ concentration will also be $0.015$ $mol/{dm}^3$

So, $pH=-log\left[H_3{O}^{+}\right]$

$\Rightarrow pH=-log\left[0.015\right]$

$\Rightarrow pH=1.82$

$1.82+pOH=14$

$\Rightarrow pOH=14-1.82=12.18$

If $pOH=-log\left[{OH}^{-}\right]$

Then ${10}^{-pOH}=\left[{OH}^{-}\right]$

Hence ${10}^{-\left(12.18\right)}=\left[{OH}^{-}\right]$

$\Rightarrow \left[{OH}^{-}\right]={10}^{-\left(12.18\right)}\approx 6.61\times {10}^{-13}mol/{dm}^3$