Question

A 20.0-mL sample of 0.150 MKOH is titrated with 0.125 MHClO4 solution. Calculate the pHafter the following volumes of acid have been added.

Part A

20.0 mL

Express your answer using two decimal places.

pH =

______

Part B

23.5 mL

Express your answer using two decimal places.

pH =

______

Part C

24.0 mL

Express your answer using two decimal places.

pH = ______

Part D

26.5 mL

Express your answer using two decimal places.

pH =

________

Part E

32.0 mL

Express your answer using two decimal places.

pH = _______

Answer 1

millimoles of KOH = 20 x 0.15 = 3.0

A) millimoles of HClO4 added = 20 x 0.125 = 2.5

3.0 - 2.5 = 0.5 millimoles NaOH left

[NaOH] = 0.5 / 40 = 0.0125 M

pOH = - log [OH^-]

pOH = - log [0.0125]

pOH = 1.903

pH = 14 - 1.903

pH = 12.09

B) millimoles of HClO4 = 23.5 x 0.125 = 2.937

3 - 2.937 = 0.063 millimoles NaOH left

[NaOH] = 0.063 / 43.5 = 0.00145 M

pOH = -log [OH^-]

pOH = - log [0.00145]

pOH = 2.84

pH = 14 - 2.84

pH = 11.16

C ) millimoles of HClO4 = 24 x 0.125 = 3.0

at equivalence point

pH = 7.0

D) millimoles of HClO4 = 26.5 x 0.125 = 3.312

3 - 3.312 = 0.312 millimoles HClO4 left

[HClO4] = 0.314 / 46.5 = 0.0067 M

pH = - log [H⁺]

pH = - log [0.0067]

pH = 2.17

E) millimoles of HClO4 = 32 x 0.125 = 4.0

4 - 3 = 1.0 millimoles HClO4 left

[HClO4] = 1 / 52 = 0.019 M

pH = - log [H⁺]

pH = - log [0.019]

pH = 1.72