Question

${10}^{-2}$ mol of $NaOH$ was added to $10$ liter of water. What will be the change in $pH$?

Answer 1

As $pH$ of water $=7$

Now, 10${}^{-2}$ mole of $NaOH$ is added into 10 L of water.

New concentration of the solution is $=\frac{{10}^{-2}mol}{10L}={10}^{-3}$ $mol/L$

$\left[O{H}^{-}\right]={10}^{-3}$ $mol/L$

$pOH=3$

$pH=14-pOH=11$

Then change in $pH$ is $=11-7=4$