Question

A titration of 25.0 mL of a solution of sodium hydroxide required 40.0 mL of 0.114M HCl to reach the equivalence point.

a)         Determine the pH of the original solution (0EP).

b)         Determine the pH of the solution at half-way to the equivalence point (1/2 EP).

c)         Determine the pH of the titration solution at the equivalence point (EP).

Answer 1

HCl + NaOH --------------> NaCl + H2O

a) calculate the initial Molarity of NaOH to find initial pH

millimoles of HCl = 40.0 x 0.114 = 4.56

4.56 millimoles NaOH must be present.

4.56 / 25 = 0.1824 M NaOH present

as NaOH is strong base [NaOH] = [OH^-] = 0.1824 M

pOH = - log [OH^-]

pOH = - log [0.1824]

pOH = 0.74

pH = 14 - 0.74

pH = 13.26

b) half way of equivalence point meand 50% NaOH must be consumed

4.56 / 2 = 2.28 millimoles NaOH must be consumed

2.28 = 0.114 x V

V = 20 mL HCl

total volume = 25 + 20 = 45 mL

[NaOH] left = 2.28 / 45 = 0.051 M

pOH = - log [OH^-]

pOH = - log [0.051]

pOH = 1.29

pH = 14 - 1.29

pH = 12.71

c) as it is strong acid and strong base titration.

at equivalence point solution will be neitral

pH = 7.0