Question

40.00 mL of 0.10 M KOH solution is titrated by 30.00 mL of 0.10 M HCl (adding HCl to KOH). What is the pH of the solution in the Erlenmeyer flask after the titration? Assume that the volumes of the solutions are additive.

Answer 1

Answer:

$pH=12.16$

Explanation:

The alkaline solution:

40 ml of 0.1 M KOH

$n_{KOH}=\frac{1 L}{1000mL}*40mL*0.1M=0.004 mol$

The acid solution:

30 ml of 0.1 M HCl

$n_{HCl}=\frac{1 L}{1000mL}*30mL*0.1M=0.003 mol$

The neutralization reaction:

$HCl + KOH \longrightarrow KCl + H_2O$

For 1 mol of HCl, 1 mol of KOH is consumed. If 0.003 mol of HCl are added, 0.003 mol of KOH reacted.

After titration:

$n_{KOH}=0.001 mol$

$n_{HCl}= 0 mol$

Concentration (don't forget to add the volumes)

$n_{KOH}=\frac{1 1000mL}{1L *70mL}*0.001 mol=0.0143 mol$

Calcualtion of the pOH:

$pOH=-log([OH^-])=-log(0.0143)=1.84$

For the pH:

$pH=14-pOH=14-184=12.16$