A 20.0-mL sample of 0.25 M HNO3 is titrated with 0.15 M Na OH. What is the pH of the solution after 30.0 mL of

Question

of NaOH have been added to the acid?

a. 2.00

b. 1.60

c. 1.05

d. 1.00

e. none of the above

Answer 1

Moles HNO3 = 0.020 L x 0.25 =0.0050

Moles NaOH = 0.030 L x 0.15 = 0.0045

Moles HNO3 in excess = 0.0050 - 0.0045 =0.00050

total volume = 0.050 L

concentration H+ = 0.00050 / 0.050 = 0.0100 M

pH = 2.00