Question

Ammonium chloride is dissolved in 250 ml solution of amonia (0.036 M) to make a buffer system. Assume a neglogable change in volume after addition of the salt.(Kb=1.8x10^-5).

a)Determine what mass of ammonium chloride is needed for maximum buffer capacity,

b) determine the pH of the solution afetr 0.0020 mol of cessium hydroxide is added to the buffer solution.(no change in volume)

c) Is this a good buffer solution?

d)Determine what mass of ammonium chloride is needed to make a buffer of pH 8.00?

Answer 1

a) The mximum buffer capacity is obtained when

[ Base ] = [ Conjucate acid ]

Therfore

[ NH4+ ] must be 0.036M

No of mole of NH4+ required 250ml = (0.036mol/1000ml)×250ml = 0.009mol

Molar mass of NH4Cl = 53.49g/mol

Mass of NH4Cl required = 0.009mol × 53.49g/mol = 0.4814g

So, 0.4814g NH4Cl salt must be added

b) CsOH react with NH4+ to form NH3

OH- + NH4+ -------> NH3 + H2O

0.0020 mol of CsOH react with 0.0020 mole of NH4+ to form 0.0020 mole of NH3

After addition

No of mole of NH3 = 0.0090 + 0.0020 = 0.0110

No of mole of NH4+ = 0.0090 - 0.0020 = 0.0070

No volume change , so

[NH3] = (0.0110mol/250ml)×1000ml =0.044M

[ NH4+ ] = (0.0070mol/250ml)×1000ml =0.0280M

Henderson - Hasselbalch equation is

pOH = pKb + log([Conjucate acid ]/[ base ])

= 4.75 + log(0.028M/0.044M)

= 4.75 - 0.20

= 4.55

pH = 14-pOH

= 14 - 4.55

= 9.45

c) This is not good buffer because buffer capacity is low

before adding CsOH , pH = 9.25

After adding CsOH , pH = 9.45

0.20 variation by 0.0020 mol of base is indicating good buffer

Buffer capacity inceased by increasing moles of NH3 and NH4+

d) pH = 8.00

pOH = 14 - 8.00 = 6

According to Henderson equation

6 = 4.75 + log ([NH4+]/[NH3])

log([NH4+ ]/[NH3]) = 1.25

[NH4+]/[NH3] = 1×10^1.25=17.78

No of mole NH4+ = 17.78 × No of mole of NH3

No of mole of NH4+ = 17.78 ×0.009 = 0.16002

Mass of NH4+ = 0.16002 ×53.49 =8.5595g

So, 8.5595g of NH4Cl must be added