# What should be the pH after 33 ml of 0.1 M NaOH was added to 40 ml of 0.05 M HCl?

Jul 3, 2017

12.25

#### Explanation:

$33 m l \left(0.10 M N a O H\right) + 40 m l \left(0.05 M H C l\right)$ => 1:1 Rxn Ratio

=> $0.033 \left(0.10\right) \text{mole" NaOH + 0.040(0.05)"mole} H C l$

=> $0.0033 \text{mole" NaOH + 0.0020 "mole} H C l$

=>$\left(0.0033 - 0.0020\right) \text{mole". "of} . N a O H$ in excess

=> $0.0013$ mole $N a O H$ remains in excess in 73 ml of solution.

=> ${\left[N a O H\right]}_{e x c e s s}$ = [OH^-]$= \left(\frac{0.0013 \text{mole} O {H}^{-}}{0.073 L i t e r S o l u t i o n}\right)$

= $0.017 M \left(O {H}^{-}\right)$

=> $p O H = - \log \left[O {H}^{-}\right] = - \log \left(0.017\right) = - \left(- 1.75\right) = 1.75$

=> $p H = 14 - p O H = 14 - 1.75 = 12.25$