Question

# 10 mL of 0.10 M HCl is being titrated with 0.10 M NaOH. What would the...

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

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

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