Question

A 50.0 mL solution of 0.167 M KOH is titrated with 0.334 M HCl . Calculate the pH of the solution after the addition of each of the given amounts of HCl .

0.00 mL pH = 7.00 mL pH = 12.5 mL pH = 19.0 mL pH = 24.0 mL pH = 25.0 mL pH = 26.0 mL pH = 31.0 mL pH =

Answer 1

Millimole of KOH in solution = Molarity x volume in mL

=0.167 M x 50 mL = 8.35 millimole

Balance equation is

KOH ( aq) + HCl ( aq) KCl ( aq) + H2O ( l )

Ionic reaction is

H⁺ ( aq) + OH^​​​- ( aq) H2O ( l )

Nature of solution will depend on concentration of proton or hydroxyl ion and pH will be determined accordingly.

1) 0 mL

[ OH^{​​​​​-} ] = 0.167 M

P^{​​​​​​OH =} - log [ OH ^-] = - log ( 0.167 )= 0.78

P^{​​​​​​H} = 14 -0.78 = 13.22

2) millimole of H⁺ = 0.334 M x 7 mL = 2.338 millimole

[ OH^{​​​​​-}] =( 8.35 -2.338)/ ( 50+ 7 ) mL = 0.1054 M

P ^OH = - log ( 0.1054 ) = 0.98

P ^H = 14 - 0.98 = 13.02

3)12.5 mL

[ OH ^- ] = ( 8.35 - 4.175 ) millimole / ( 50 + 12.5 ) mL

P^{​​​​​H =} 14 - 1.18 = 12.82

4) 19 mL

[ OH ^- ] = ( 8.35 - 6.35 ) millimole /( 50 + 19 ) mL

P ^H = 12.46

5) 24 mL

[ OH ^- ] = ( 8.35 -8.016 ) millimole / ( 50 + 24 ) mL

P ^{H =} 11.65

6 ) 25 mL

[ H ⁺ ​​​​​​] = [ OH ^- ]

P ^{H =} 7

7) 26 mL

[ H ⁺ ] = ( 8.684 - 8.35 ) millimole / ( 50+26 ) mL

P ^{​​​​​​H =} 2.35

8 )31 mL

[ H ⁺ ] =( 10.35 - 8.35 ) millimole /( 50 + 31 ) mL

P ^H = 1.6