In: Chemistry
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 .
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
POH = - log [ OH -] = - log ( 0.167 )= 0.78
PH = 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
PH = 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