Question

Consider the titration of a 26.9 −mL sample of 0.115 M RbOH with 0.100 M HCl. Determine each of the following. C. the pH at 5.1 mL of added acid? D. the pH at the equivalence point? E. the pH after adding 4.1 mL of acid beyond the equivalence point?

Answer 1

C)

millimoles of Base = 26.9 x 0.115 = 3.094

millimoles of acid = 0.1 x 5.1 = 0.51

base millimoles > acid millimoles

[OH-] = 3.094 - 0.51 / (26.9 + 5.1)

= 0.081 M

pOH = -log [OH-] = -log (0.081) = 1.09

pH + pOH = 14

pH = 12.91

D)

at equivalence point : pH = 7

because it is strong acid and strong base titration

E)

at equivalence point volume of HCl:

26.9 x 0.115 = 0.1 x V

V = 30.94 mL

now the volume added = 30.9 + 4.1 = 35.0 mL

acid miilimoles = 35 x 0.1 = 3.5

acid > base

[H+] = 3.5 - 3.09 / (5.1 + 35)

= 0.010 M

pH = -log [H+] = -log (0.010)

pH = 2.0