Question

10 mL of 0.10 M HCl is being titrated with 0.10 M NaOH. What would the pH of the solution be after addition of a) 1.0 mL of NaOH and b) 9.0 mL of NaOH

Answer 1

Both HCl and NaOH undergo ionization completely since they are strong acid and base respectively.

The reaction between NaOH and HCl can be represented as

NaOH+ HCl----> NaCl+ H2O   (1 mole of HCl reacts with 1 mole of NaOH)

moles in 0.1M containing 10ml =0.001 moles

a) moles of NaOH in 0.1M and 1 ml= 0.0001 moles

HCl is excess and all the NaOH gets consumed

So moles of HCl remaining =0.001-0.0001=0.0009

Total volume after mixing = 10 (HCl)+ 1ml( NaOH)= 11ml

Molarity of HCl= 0.0009*1000/11=0.0818

HCl----> H+ Cl-

pH= -log(0.0818)=1.087

b) Moles of NaOH in 0.1M and 9ml =9*0.1/1000=0.0009

HCl is excess and all the NaLOH reacts and HCl remaining =0.001-0.0009=0.0001

Volume after mixing = 10+9=19

Molarity= 0.0001*1000/19=0.005263

pH= -log(0.005263)=2.278