Question

A 20.0 mL sample of 0.200 MHBr solution is titrated with 0.200 MNaOH solution. Calculate the pH of the solution after the following volumes of base have been added.

Part A

16.0 mL

Part B

19.8 mL

Part C

20.0 mL

Part D

20.1 mL

Part E

36.0 mL

Answer 1

A)

moles of acid = 0.2 M * 20 mL = 4 mmol

moles of base added = 0.2 M * 16 mL = 3.2 mmol

3.2 mmol of each will react with each other

after reaction,

moles of acid remaining = 4-3.2 = 0.8 mmol

total volume = 20 +16 = 36 mL

[H+] = 0.8/36=0.022 M

pH = -log [H+]

= -log (0.022)

= 1.65

　

B)

moles of acid = 0.2 M * 20 mL = 4 mmol

moles of base added = 0.2 M * 19.8 mL = 3.96 mmol

3.96 mmol of each will react with each other

after reaction,

moles of acid remaining = 4-3.96= 0.04 mmol

total volume = 20 +19.8 = 39.8 mL

[H+] = 0.04/39.8=1*10^-3 M

pH = -log [H+]

= -log (1*10^-3)

= 3

C)

equal moles of acid and base has been added

Hence pH = 7

D)

moles of acid = 0.2 M * 20 mL = 4 mmol

moles of base added = 0.2 M * 20.1 mL = 4.02 mmol

4 mmol of each will react with each other

after reaction,

moles of Base remaining = 4.02 -4= 0.02 mmol

total volume = 20 +20.1 = 40.1 mL

[OH-] = 0.02/40.1=4.99*10^-4 M

pOH = -log [OH-]

= -log (4.99*10^-4)

= 3.3

pH= 14 - 3.3

= 10.7

I am allowed only 4 parts at a time