Question

Part 1:

You wish to create a buffer with the pH equal to the pKa. If the following combinations were dissolved in 1.0 L of water, which combination(s) would accomplish that goal?

1) 1.0 mole of HF and 1.0 mol of NaF

2) 1.0 mole of HF and 1.0 mol of NaOH

3) 1.0 mole of NaF and 1.0 mol of HCl

4) 1.0 mole HF and 0.5 mol NaOH

A

1, 2, 3, and 4

B

1 only

C

1 and 4 only

D

None of these

E

2, 3 and 4 only

Part 2:

Nitrous acid has a Ka of 4.0E-4. In 1.00 L of solution, 0.700 moles of nitrous acid (HNO2) are added to 0.265 moles of NaOH. What describes the end result?

A

A solution of weak acid in water

B

A buffer solution in water

C

A solution of strong base in water

D

None of these

E

A solution of a strong acid in water

Part 3:

Nitrous acid has a Ka of 4.0E-4. in 1.00 L of solution, 0.700 moles of nitrous acid (HNO2) are added to 0.265 moles of NaOH. What is the final pH?

Answer 1

Part-1)

All the reagent like HF, HCl, NaOH, NaF are strong electrolytes (acids/bases/salts) and hence no combination will result in Buffer formation.

To form a buffer solution one need to use a weak acid + corresponding conjugate base (for acidic buffer) or

base + corresponding conjugate acid (for basic buffer).

Answer option : (D) None of these.

Part 2)

A pertinent neutralization reaction between Nitrous acid (HNO2) and NaOH is

HNO₂ + NaOH NaNO₂ + H₂O

There is 1:1 Acid:base equivalence.

0.265 moles of NaOH will neutralize only 0.256 moles of the 0.700 moles HNO2 and

0.700 - 0.256 = 0.444 moles.

It means on mixing we have [NaNO2] : [HNO2] = 0.256 : 0.444

This forms a pair of weak acid + conjugate base which forms a Buffer in water.

Answer option : (B)

Part 3)

Ka of HNO2 = 4.0 x 10^-4.

pKa = -log(4.0 x 10^-4) = 3.36

On adding 0.256 moles of NaOH to 0.7 M HNO2 we get (we calculated it in Part 2)

[NaNO2] : [HNO2] = 0.256 : 0.444

By Henderson-Hasselbalch equation we get,

pH = pKa + log ([conjugate base or salt]/[Acid])

pH = pKa + log ([NaNO2]/[HNO2])

pH = pKa + log (moles of NaNO2/ moles of HNO2)

pH = 3.36 + log (0.256 / 0.444)

pH = 3.36 + (-0.24)

pH = 3.12

pH of resultant buffer is 3.12

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