Question

What is the pH (to nearest 0.01 pH unit) of a solution prepared by mixing 77.0 mL of 0.0699 M NaOH and 77.0 mL of 0.0362 M Ba(OH)2?

pH=________

What is the pH (to nearest 0.01 pH unit) of a solution prepared by mixing 65.0 mL of 0.736 M HCl and 32.0 mL of 1.869 M NaOH?

pH =________

Answer 1

total OH ions

OH from NaOH = MV = 0.0699*77 = 5.5748

oH from Ba(OH) = 2*MV = 2*0.0362*77 = 5.3823

total OH = 5.3823+5.5748 = 10.9571

total V= 77+77 = 154

[OH-] = 10.9571/154 = 0.07115 M

pOH = -log(0.07115)

pH = 14-1.1478 = 12.85

2

mmol of acid = MV = 65*0.736 = 47.84

mmol base = MV = 1.869*32 = 59.808

mmol of base left = 59.808 -47.84 = 11.968 mmol of base

then

VT = V1+V2 = 65+32 = 97

[OH-] = 11.968/97 = 11.968/97

pOH =-log( 11.968/97) = 0.90875

pH = 14-0.90875 = 13.09125