Question

Calculate the pH of a solution from the titration of 25.0 mL of 0.125 M HCl with 20.0 mL of 0.100 M KOH

Answer 1

we have:

Molarity of HCl = 0.125 M

Volume of HCl = 25 mL

Molarity of KOH = 0.1 M

Volume of KOH = 20 mL

mol of HCl = Molarity of HCl * Volume of HCl

mol of HCl = 0.125 M * 25 mL = 3.125 mmol

mol of KOH = Molarity of KOH * Volume of KOH

mol of KOH = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 3.125 mmol

mol(KOH) = 2 mmol

2 mmol of both will react

remaining mol of HCl = 1.125 mmol

Total volume = 45.0 mL

[H+]= mol of acid remaining / volume

[H+] = 1.125 mmol/45.0 mL

= 0.025 M

we have below equation to be used:

pH = -log [H+]

= -log (2.5*10^-2)

= 1.60

Answer: 1.60