Question

How many moles of NH3 are required in a 1.00L solution to create a buffer with a pH of 9.08 if the solution contains 0.78 moles of NH4Cl? Kb NH3 = 1.8 x 10-5?

How many moles of HClO are required in a 1.00L solution to create a buffer with a pH of 7.05 if the solution contains 0.32 moles of NaClO? K_aHClO = 3.0 x 10^-8?

What is the pH after 0.033 moles of HCl is added to a 100.0mL buffer solution containing 0.65M HOBr and 0.55M NaOBr ? K_{a HOBr} = 2.5 x 10^-9

Answer 1

1.

Buffer problems are best solved using the Henderson-Hasselbalch equation:

pOH = pKb + log ( [salt] / [base] ),

where salt is NH4Cl and base is NH3 part of the buffer.

Since the volume is 1 L, the equation can also be written as

pOH = pKb + log (moles NH4Cl / moles NH3)

pH = 9.08

So, pOH = 14 – 9.08 = 4.92

Kb for NH3 = 1.8 x 10^-5;

pKb = -log Kb = -log (1.8 x 10^-5) = 4.75

So, 4.92 = 4.75 + log (0.78 / moles NH3)

0.17 = log (0.78 / moles NH3)

(0.78 / moles NH3) = 10^0.17

(0.78 / moles NH3) = 1.48

moles NH3 = 0.78 / 1.48 = 0.53 moles

2.

Buffer problems are best solved using the Henderson-Hasselbalch equation:

pH = pKa + log ( [salt] / [acid] ),

where salt is NaClO and acid is HClO part of the buffer.

Since the volume is 1 L, the equation can also be written as

pH = pKa + log (moles NaClO / moles HClO)

pH = 7.05

Ka for HClO = 3.0 x 10^-8;

pKa = -log Ka = -log (3.0 x 10^-8) = 7.52

So, 7.05 = 7.52 + log (0.32 / moles HClO)

- 0.47 = log (0.32 / moles HClO)

(0.32 / moles HClO) = 10^-0.47

(0.32 / moles HClO) = 0.34

moles HClO = 0.32 / 0.34 = 0.94 moles

3.

Buffer problems are best solved using the Henderson-Hasselbalch equation:

pH = pKa + log ([salt] / [acid]),

where salt is NaOBr and acid is HOBr, part of the buffer.

Ka for HOBr = 2.5 x 10^-9;

pKa = -log Ka = -log (2.5 x 10^-9) = 8.60

So,

pH_initial = 8.60 + log (0.55 / 0.65)

     = 8.60 + log (0.846)

     = 8.60 – 0.07

     = 8.53

Molarity = Moles/Liter

Moles = Molariy x Liter

Initial moles HOBr = (0.65 M) (0.100 L) = 0.065 moles acid
initial moles NaOBr = (0.55 M) (0.100 L) = 0.055 moles salt

When HCl is added, we convert some NaOBr to HOBr. So, moles of NaOBr decreases and moles of HOBr increases by the moles of HCl added.

Now, moles of acid and salt after addition of HCl

Initial moles acid = 0.065 moles + 0.033 moles = 0.098 moles
initial moles salt = 0.055 moles – 0.033 moles = 0.022 moles

Concentration terms are

[HOBr] = 0.098 moles = 0.1 L = 0.98 M
[NaOBr] = 0.022 moles / 0.1 L = 0.22 M

So,

pH_final = 8.60 + log (0.22 / 0.98)

          = 8.60 + log (0.224)

          = 8.60 – 0.65

          = 7.95

Change in pH = pH_initial - pH_final

                      = 8.53 – 7.95

                      = 0.58