How many moles of NH4Cl must be added to 100 mL of 0.1 M NH4OH solution to preve
ID: 907977 • Letter: H
Question
How many moles of NH4Cl must be added to 100 mL of 0.1 M NH4OH solution to prevent precipitation of Mn(OH)2 when this solution is added to 100 mL of a 0.002 M solution of MnCl2? Assume no change involume on addition of NH4Cl. How many moles of NH4Cl must be added to 100 mL of 0.1 M NH4OH solution to prevent precipitation of Mn(OH)2 when this solution is added to 100 mL of a 0.002 M solution of MnCl2? Assume no change involume on addition of NH4Cl. How many moles of NH4Cl must be added to 100 mL of 0.1 M NH4OH solution to prevent precipitation of Mn(OH)2 when this solution is added to 100 mL of a 0.002 M solution of MnCl2? Assume no change involume on addition of NH4Cl.Explanation / Answer
(i) Before addition of NH4Cl
100 ml of 0.1 M NH4OH
100 ml of 0.002 M MnCl2
Concentration of mixture
[NH4OH] = 0.1*100/(100+100)=0.05 M
[MnCl2] = 0.002*100/(100+100)=0.001 M
Solubility equilibria of Mn(OH)2
Mn(OH)2(s) = Mn(+2)(aq) + 2OH-(aq);
Ksp = [Mn(+2)]*[OH-]^2 = 1.6E-13;
[Mn(+2)]= [MnCl2] = 0.001 M
For precipipitation, Value of [OH-] = (Ksp/[Mn(+2+])^0.5 = ([1.6E-13]/[0.001])^0.5 = 1.265E-5
pOH = -Log([OH-]) = 4.9
pH = 14 - pOH = 9.1
(ii) After addition of NH4Cl with NH4OH buffer solution will be created having pH = pKa + Log([Base]/[Acid])
pKa of ammonia = 9.25
pH = pKa + Log([NH4OH]/[NH4Cl]) =
Log([NH4OH]/[NH4Cl]) = pH - pKa = 9.1-9.25 = -0.15
[NH4OH]/[NH4Cl] = 0.71
[NH4Cl] = 0.71*[NH4OH] = 0.71*0.05 = 0.035 M
Total volume of solution, V = 100+100 = 200ml = 0.2 L
Moles of NH4Cl = 0.035*0.2 = 0.007 moles