Question

A 50.0 ml sample of 0.100 M Ba (OH)2 is treated with 0.100 M HNO3. Calculate the ph after the addition of the following volumes of acid.
a). 25 ml of HNO3
b). 50 ml of HNzo3
c). 75 ml of HNO3
d). 100 ml of HNO3
e). 125 ml of HNO3

Answer 1

each Ba(OH)2 will produce 2 OH- ions

each HNO3 will produce 1 H+ ion

a) No. of mmols of Ba(OH)2 = 2 x 0.100 x 50 = 10 mmols

No. of mmols of HNO3 = 1 x 0.100 x 25 = 2.5 mmol

Here, OH- conc. is excess So, excess OH = 10 - 2.5 = 7.5 mmol

[OH-] = 7.5 / (25 + 50) = 0.1

pOH = - log[0.1] = 1; So, pH = 14 - 1 = 13

b)

No. of mmols of HNO3 = 1 x 0.100 x 50 = 5.0 mmol

Here, OH- conc. is excess So, excess OH = 10 - 5.0 = 5.0 mmol

[OH-] = 5.0 / (50 + 50) = 0.05

pOH = - log[0.05] = 1.3; So, pH = 14 - 1.3 = 12.7

c)

No. of mmols of HNO3 = 1 x 0.100 x 75 = 7.5 mmol

Here, OH- conc. is excess So, excess OH = 10 - 7.5 = 2.5 mmol

[OH-] = 2.5 / (75 + 50) = 0.02

pOH = - log[0.02] = 1.7; So, pH = 14 - 1.7 = 12.3

d)

No. of mmols of HNO3 = 1 x 0.100 x 100 = 10.0 mmol

Here, OH- conc. is excess So, excess OH = 10 - 10 = 0 mmol

This is a neutral solution; Have pH = pOH = 7

e)

No. of mmols of HNO3 = 1 x 0.100 x 125 = 12.5 mmol

Here, H conc. is excess So, excess H+ = 12.5 - 10 = 2.5 mmol

[H+] = 2.5 / (125 + 50) = 0.01428

pH = - log[0.01428] = 1.845