Question

A 15.0-mL sample of 0.100 M Ba(OH)2 is titrated with 0.125 MHCl.

Calculate the pH for different points throughout the titration curve and make a sketch of the curve.

Answer 1

Let us take 4 points those are 0 ml, 10 ml , 30 ml and finally 50 ml

1) at 0 ml of 0.125 MHCl:

Ba(OH)2 will dissociate as below

Ba(OH)2 = Ba2+ + 2OH-

That menas one mole of Ba(OH)2 will give us 2 moles of OH-

Hence [OH-] = 0.2 M

pOH = -log[OH-] = -log0.2 = 0.699

pH = 14-0.699 = 13.301

2) at 10 ml of 0.125 MHCl:

Ba(OH)2 + 2HCl = BaCl2 + 2H2O

No. of moles of Ba(OH)2 = 0.1 x 0.015 = 0.0015 moles

No. of moles of acid = 0.125 x 0.010 =   0.00125 moles

After reaction the remaining Ba(OH)2 = 0.0015 - 0.00125/2 = 0.000875 moles

Total volume = 25 ml, hence concentration = 0.000875 moles/0.025L = 0.035 M

Hence [OH-] = 0.07, pOH = 1.15

pH = 12.85

3) Similarly if calculated for 30 ml we will get 0.000375 moles of acid, total volume = 45 ml

     Concentration of HCl = 0.00833 M

      pH = 2.08

4) Similarly if calculated for 50 ml we will get 0.001625 moles of acid, total volume = 65 ml

     Concentration of HCl = 0.025 M

      pH = 1.6

Now Sketch the curve by using the points (0, 13.3) , ( 10, 12.85) , (25, 2.08), (50, 1.6)