Question

Calculate the pH of the resulting solution if 29.0 mL of 0.290 M HCl(aq) is added to

(a) 34.0 mL of 0.290 M NaOH(aq).

pH:?

(b) 39.0 mL of 0.340 M NaOH(aq).

pH:?

Answer 1

millimoles of HCl = 29 x 0.290 = 8.41

(a) 34.0 mL of 0.290 M NaOH(aq).

millimoles of NaOH = 34 x 0.290 = 9.86

base millimoles > acid millimoles

[OH-] = 9.86 - 8.41 /(34+29) = 0.023 M

pOH = -log[OH-] = -log (0.023) = 1.64

pH + pOH = 14

pH = 12.36

(b) 39.0 mL of 0.340 M NaOH(aq).

millimoles of NaOH = 39 x 0.340 = 13.26

base millimoles > acid millimoles

[OH-] = 13.26 - 8.41 /(39+29) = 0.0713 M

pOH = -log[OH-] = -log (0.0713) = 1.15

pH + pOH = 14

pH = 12.85