Question

${10}^{-3}$ mole of $KOH$ dissolved in $100L$ of water. Calculate the $pH$ of the solution.

Answer 1

Number of moles = ${10}^{-3}$ and the volume of the solution is 100L

$\therefore Molarity=\frac{moles}{volumeofsolutionin{}^{\prime }{L}^{\prime }}=\frac{{10}^{-3}}{100}={10}^{-5}mol/litre$

A strong acid like KOH undergoes dissociation as follows :-

$KOH\to K^{+}+O{H}^{-}$

initial concentration ${10}^{-5}$ $0$ $0$

concentration at equillibrium $0$ ${10}^{-5}$ ${10}^{-5}$

Thus the concentration of hydroxyl ions $\left(O{H}^{-}\right)$ at equilibrium is ${10}^{-5}mol/litre$

$\therefore pOH=-lo{g}_{10}\left[O{H}^{-}\right]=-lo{g}_{10}\left[{10}^{-5}\right]=5$

$\because pOH=5$

$\therefore pH=14-pOH$

$=14-5$

$=9$

Hence, the pH of the solution is 9.