What should be the pH after 33 ml of 0.1 M NaOH was added to 40 ml of 0.05 M HCl?

1 Answer
Jul 3, 2017

12.25

Explanation:

33ml(0.10MNaOH)+40ml(0.05MHCl) => 1:1 Rxn Ratio

=> 0.033(0.10)moleNaOH+0.040(0.05)moleHCl

=> 0.0033moleNaOH+0.0020moleHCl

=>(0.00330.0020)mole.of.NaOH in excess

=> 0.0013 mole NaOH remains in excess in 73 ml of solution.

=> [NaOH]excess = [OH^-]=(0.0013moleOH0.073LiterSolution)

= 0.017M(OH)

=> pOH=log[OH]=log(0.017)=(1.75)=1.75

=> pH=14pOH=141.75=12.25