Question

A 40.0 mL sample of 0.150 M Ba(OH)2 is titrated with 0.400 M HNO3. Calculate the pH after the addition of the following volumes of acid.

a) 0.0 mL
b) 15.0 mL
c) At the equivalence point

d) 40.0 mL

Answer 1

a)

Ba(OH)2----> Ba⁺² + 2OH-

The reaction between Ba(OH)₂ + 2HNO3---> Ba(NO3₎₂+2H2O

0.15M Ba(OH)2 correspond to 2*0.15M OH= 0.3M

pOH= -log (0.3)=0.53

pH= 14-pOH= 14-0.53=13.47

b. moles of Ba(OH)2 in 40ml of 0.150= 0.150*40./1000=0.006 moles

OH- = 2*0.006= 0.012 moles

moles of HNO3= 0.4*15/1000= 0.006

Excess of OH-= 0.012-0.006 =0.006

Volume after mixing =40+15= 55ml =55/1000L

Concentration of OH-= 0.006*1000/55=0.109 M

pOH= -log (0.109)= 0.9626

pH= 14-0.9626=13.0374

c) moles of Ba(OH)2= 0.006

Because both are storng acid and strong base, pH at equivalene point =7

d) when 40ml 0.4M HNO3 is added= 0.4*40/1000= 0.016 moles

0.006 moles of Ba(OH)2 requires 0.012 moles of HNO3. So moles of HNO3 remaining= 0.016-0.012=0.004

Volume after mixing= 40+40= 80ml

Concentration of HNO3= 0.004*1000/80=0.05

HNO3----> H+ + NO3-

H+ =0.05

pH= -log (0.05)=1.3