Question

Explain how you would make 1.0 L of a 1.05 M acetate buffer if you only have

sodium acetate, DI H2O and 6 M HCl available? PH is 4.6

The pKa of ammonium (NH4+) is 9.24.

a. What is the predominant form (ammonium or ammonia) of this

chemical when the pH of the solution is 7.85?

b. One liter of solution is made using 0.25 moles of NH4Cl and 0.07 moles

of NH3. What is the pH of this solution? Recall that the Henderson-

Hasselbalch equation can be written in the form pKa = pH –

log([base]/[acid])

c. You have NaOH and HCl at your disposal. Which one would you use,

and how many moles would you need, in order to titrate the solution to

pH 9.86? Ignore any change in volume.

Answer 1

1) Preparation of 1.0L of 1.05M acetate buffer:

molar mass of sodium acetate=82.034g/mol

So grams of sodium acetate to be taken to prepare 1.05 mol/L of buffer=1.05 mol*82.034 g/mol=86.136 g per L

86.136 g of sodium acetate(solid) is to be taken and dissolved in 1 L DI to prepare 1.05M sodium acetate solution.Then,6M HCl is to be added to neutralize sodium acetate partly,to form buffer.

Buffer sytem equation,

CH3COONa ----->CH3COO^- +Na⁺

CH3COO^- +H2O <--->CH3COOH +OH^-

pH=4.6

pka of acetate/acetic acid =4.75

Using Henderson-Hasselbach equation,

pH=pKa+ log (base/acid)

pH=pka+log [acetate]/[acetic acid]

4.6=4.76+log[acetate]/[acetic acid]

-0.16=log[acetate]/[acetic acid]

[acetate]/[acetic acid]=10^-0.16=0.692

HCl neutralizes sodium acetate to give acetic acid in the ratio [acetate]/[acetic acid]=0.692:1

HCl+CH3COONa ---->CH3COOH +NaCl

So, [acetate] + [acetic acid]=1.05M

Thus, [acetate]=0.692/(1+0.692)* total concentration=0.429M

[acetic acid]=1.05M-0.0429M=0.620M

moles of acetic acid in 1L=0.620mol/L*1L=0.620 mol

volume of 6M HCl to be added=0.620 mol/6 mol/L=0.103 L=0.103L*1000ml/L=103 ml

So,103 ml of HCl is to be added in the prepared solution to form a buffer 1.05 M of 1L volume