Question

Calculate the ph of a solution which contain 9.9ml of 1m hcl and 100 ml of 0.1 m naoh

Answer 1

NaOH reacts with HCl :

NaOH + HCl → NaCl + H2O

1 mol NaOH reacts with 1 mol HCl

Mol HCl in 9.9mL of 1.0M solution = 9.9/1000*1.0 = 0.0099 mol HCl

Mol NaOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol NaOH

On mixing the 0.0099 mol HCl will be neutralised and 0.01 - 0.0099 = 1*10^-4 mol NaOH dissolved in 109.9mL solution

Molarity of NaOH solution = (1*10^-4) / 0.1099 = 9.099*10^-4M

pOH = -log ( 9.099*10^-4)

pOH = 3.04

pH = 14.00 - pOH

pH = 14.00 - 3.04

pH = 10.96