15. What is the pH of a solution made by combining 150 mL of 0.125 MHCl and 150 mL of 0.120 M Ba(OH)2 at 25°C? A) 12.760 B) 1.240 C) 2.602 D) 11.398 Solution 15) Given: M(H+) = 0.125 M V(H+) = 150 mL M(OH-) = 2*[Ba(OH)2] = 2*0.120 M = 0.240 M V(OH-) = 150 mL mol(H+) = M(H+) * V(H+) mol(H+) = 0.125 M * 150 mL = 18.75 mmol mol(OH-) = M(OH-) * V(OH-) mol(OH-) = 0.24 M * 150 mL = 36 mmol We have: mol(H+) = 18.75 mmol mol(OH-) = 36 mmol 18.75 mmol of both will react remaining mol of OH- = 17.25 mmol Total volume = 300.0 mL [OH-]= mol of base remaining / volume [OH-] = 17.25 mmol/300.0 mL = 0.0575 M use: pOH = -log [OH-] = -log (5.75*10^-2) = 1.240 use: PH = 14 - pOH = 14 - 1.240 = 12.760 Answer: A .

1. 15. What is the pH of a solution made by combining 150 mL of 0.125 MHCl and 150 mL of
0.120 M Ba(OH)2 at 25°C? A) 12.760 B) 1.240 C) 2.602 D) 11.398
Solution
15)
Given:
M(H+) = 0.125 M
V(H+) = 150 mL
M(OH-) = 2*[Ba(OH)2] = 2*0.120 M = 0.240 M
V(OH-) = 150 mL
mol(H+) = M(H+) * V(H+)
mol(H+) = 0.125 M * 150 mL = 18.75 mmol
mol(OH-) = M(OH-) * V(OH-)
mol(OH-) = 0.24 M * 150 mL = 36 mmol
We have:
mol(H+) = 18.75 mmol
mol(OH-) = 36 mmol
18.75 mmol of both will react
remaining mol of OH- = 17.25 mmol
Total volume = 300.0 mL

2. [OH-]= mol of base remaining / volume
[OH-] = 17.25 mmol/300.0 mL
= 0.0575 M
use:
pOH = -log [OH-]
= -log (5.75*10^-2)
= 1.240
use:
PH = 14 - pOH
= 14 - 1.240
= 12.760
Answer: A