Question

What is pH of 50 ml 0.05 (M) $NaOH$ and 50 ml 0.1 (N) $C{H}_{3}C{O}_{2}H$

Answer 1

NaOH reacts with $C{H}_{3}COOH$ in 1:1 molar ratio.

You are mixing equimolar amounts of the two reactants . You will produce a solution of $C{H}_{3}COONa$

Mol $C{H}_{3}COOH$ in 50.0mL of 0.1M solution $=50/1000\times 0.1=0.005$ mol $C{H}_{3}COOH$

This will produce 0.005 mol $C{H}_{3}COONa$ dissolved in 100mL = 0.1L solution

Molarity of $C{H}_{3}COONa$ solution = 0.005/0.1 = 0.05M $C{H}_{3}COONa$ solution.

$K_{a}CH3COOH=1.8\times {10}^{-5}$

$K_b={10}^{-14}/(1.8\times {10}^{-5})$

$K_b=5.56\times {10}^{-10}$

Use $K_b$ equation to calculate $\left[O{H}^{-}\right]$

$K_b=\left[O{H}^{-}{]}^2/\right[C{H}_{3}COONa]$

$5.56\times {10}^{-10}=[O{H}^{-}{]}^2/0.05$

$[O{H}^{-}{]}^2=(5.56\times {10}^{-10})\times 0.05$

$[O{H}^{-}{]}^2=2.78\times {10}^{-11}$

$\left[O{H}^{-}\right]=5.27\times {10}^{-6}$

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

$pOH=-log(5.27\times {10}^{-6})$

$pOH=5.28$

$pH=14-pOH$

$pH=14-5.28$

$pH=8.72$