A 20.0-mL sample of 0.30 M HBr is titrated with 0.15 M NaOH. What is the pH?

Question

of the solution after 40.3 mL of NaOH have been added to the acid?

a. 2.95

b. 3.13

c. 10.87

d. 11.05

e. 13.14

Answer 1

mol HBr = 0.30M * 0.020L = 0.006mol HBr = 0.006mol H3O+

#mol NaOH = 0.15M *0.0403L = 0.006045mol NaOH = 0.006045mol OH-

Limiting agent therefore is OH-, with 4.5E-5mol remaining.

#mol/L OH- = 4.5E-5/ (0.020+0.0403L) = 7.462686567E-4M

pOH = -log([OH-]) = 3.127

We know that pH + pOH = 14, so pH = 10.873

[Answer: see above]