Question

What volumes of 1.0 M NH4NO3 and 1.0 M NH3 would be required to prepare 2.00 L of a buffer

with a pH of 9.00? The Kb of NH3 is 1.8 × 10–5.

How many grams of solid NaOH would need to be added to 5.00 L of 0.50 M acetic acid to

prepare a buffer with a pH of 5.00, assuming no volume change occurs upon the addition? The Ka of

acetic acid is 1.8 × 10–5.

Answer 1

let x= volime of NH4NO3 then 2-x= volume of NH3

The reaction is NH3+HNO3----> NH4NO3

Mole of NH3 in the sample= x*1, moles of NH4NO3= 1*(2-x)

after the volume becomes 2 l

concentrations : NH3= x/2 and and NH4NO3= (2-x)/2

pH=14- pKb-+log [(concentration of NH3)/(concentration of NH4NO3)]

Kb= 1.8*10-5, pKb = 4.74

9= 14-4.74-+log (x/(2-x)

9-9.26= log {x/(2-x)}

x/(2-x)= 0.55

x= 0.55*(2-x)

1.55x= 1.1

x= 1.1/1.55 =0.71 L of NH3 and 2-0.71= 1.29 L of NH4NO3

b)

et x= gams of a NaOH to be added, mole of NaOH= x/40 moles=0.025x moles

moles of acetic acid in 5L= 0.5*5= 2.5 moles

let NaOH the limiting reactants. So moles of sodium acetate foremd= 0.25x according to

CH3COOH+ NaOH---->CH3COONa + H2O

Mole of acetic acid remaining = 2.5-0.25x

concentrations after reaction . CH3COOH= (2.5-0.25x)/5 and sodium acetate= 0.25x/2.5

pH= pKa+log [Concentration of coonjugate base]/[Concentration of acetic acid]

5= 4.74+log (0.25x/(2.5-0.25x)

0.26= log (0.25x/(2.5-0.25x)

1.82= 0.25x/(2.5-0.25x)

1.82*(2.5-0.25x)= 0.25x

4.55- 0.455x= 0.25x

4.55 = 0.705x

x= 4.55/0.705= 6.45 gms