7.66 Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl

b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2

c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

(a)MolesofH3O+=25×0.11000=.0025mol
molesofOH-=10×0.2×21000=.0040mol
Thus,excessofCH-=.0015mol
[OH-]=.001535×10-3mol/L=.0428
pOH=-log[OH]
=1.36
pH=14-1.36
=12.63
(b)MolesofH3O+=2×10×.011000=.0002mol
MolesofOH-=2×10×.011000=.0002mol

SincethereisneitheranexcessofH3O+orOH-
(c)MolesofH2O+=2×10×0.11000=.002mol
MolesofOH-=10×0.11000=0.001mol
ExcessofH3O+=.001mol
Thus,[H3O+]=.00120×10-3=10-320×10-3=0.5
pH=-log(0.05)
=1.30
Thesolutionisneutral.Hence,pH=7.