Ammonium chloride is dissolved in 250 ml solution of amonia (0.036 M) to make a
ID: 559013 • Letter: A
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?
Explanation / Answer
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
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.