How do I do a Ph question like this?

Question

The following question refers to the titration of 25.00 mL of 0.107 M HCl with 0.110 M Ba(OH)2. The reaction proceeds according to the following stoichiometry:

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

What will the pH of the titrated solution be after the addition of 20.00 mL of Ba(OH)2?

Answer 1

(25.00ml) * (0.107 M HCL) (1 M BaOH2/2M HCL)= x ml *0.110M BaOH2

x is the number of ml of BaOH that can be neutralized by the HCL is 12.159 ml round to 12.16 ml Ba(OH)2, so we will have 7.84 ml of (BaOH) that is unneutralized.

(7.84 ml)(0.110 Moles/1000ml)(2 mole OH-/1 mole BaOH2) =
number of moles of 0H- ion 0.00172 moles of OH in 45 ml water.

0.00172 mole OH/45 ml = x moles OH-/1000 ml, solve for x

0.038 molar OH- concentration of the resulting solution.

Kw = [H+] [0H-] = [H+] [ 0.038] = 10^(-14), solve for [H+]
[H+] = 2.6 * 10^(-13)
ph = - log[H+] = 12.6

Answer 2

25 ml * .107 M HCl = 2.675 mmol H+
20 ml * .11 M Ba(OH)2 = 4.40 mmol OH-

Resulting solution has 1.73 mmol OH- in 45 ml = .0384 M OH- = 2.60 M H+ = pH 12.6

In my book, that's a titration that is WAY overshot. :-)

Answer 3

no of moles of HCl added=0.107 X (25/1000)=0.002675
no of moles of Ba(OH)2 added=0.110 x (20/1000)=0.00220
By right, Ba(OH)2 and HCl with react through nuetralisation to form salt and water. But, in this case, Ba(OH)2 is given in excess.

Since BaCl2 is a nuetral salt, the pH in the resultant solution is due to the excess Ba(OH)2:
no of moles of Ba(OH)2 reacted with HCl=0.002675 / 2 =0.0013375
no of moles of Ba(OH)2 in excess: 0.00220-0.0013375=8.625 X10^-4
Ba(OH)2 -> Ba2+ + 2 OH-
no of moles of OH- = 2X8.625 X10^-4 =0.001725
[OH-]=0.001725/(45/1000)=0.03833333...
pOH=-lg[OH-]=1.4164
pH=14-pOH=14-1.4164=12.58(2dp)

-hope it helps :)