PCH241P MAYIJUNE 2017 Question 4 (a) Consider the following () The Ka value of a
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PCH241P MAYIJUNE 2017 Question 4 (a) Consider the following () The Ka value of acetic acid at 25°C has been found to be 1 8 x 105 Calculate the pKa (u) The Kb value of Hydroxylamine at 25°C has been found to be 1 1x 108 Calculate the value pko value () The pH of a solution is found to be 4 66 What are the concentrations of [H] and IOH respectively? (b) Calculate the pH of a solution which contains 600 0 cm3 of 0 300 mol L1 NH3 solution, 25 g of NHCI (solid) and 350 0 cm3 of water Kw1 x 10-14 Kb (NH3)-1 75 x 10-5 , or Ka (NH4*)-5 714 x 10.10 Molar mass NHC 53 492 g mol-1 (10) (c) Wnte the expression for the solublity product constant (Ksp) for a hypothetical species A2B3s) 1151Explanation / Answer
Part a
I) acid dissociation constant Ka = 1.8*10^-5
pKa = - log (Ka) = - log (1.8*10^-5)
= 4.74
II)
Base dissociation constant Kb = 1.1*10^-8
pKb = - log (Kb) = - log (1.1*10^-8)
= 7.96
III)
pH = 4.66
pOH = 14 - pH = 14 - 4.66 = 9.34
[H+] = 10^-pH = 10^(-4.66) = 2.187 x 10^-5 M
[OH-] = 10^-pOH = 10^(-9.34) = 4.57 x 10^-10 M
Part b
Initial moles of NH3 = molarity x volume
= 0.300 mol/L x 600 cm3 x 1L/1000cm3
= 0.180 mol
Initial moles of NH4Cl = mass/molecular weight
= 2.5g / 53.492g/mol
= 0.04674 mol
Total volume of solution = 600 + 350
= 950 cm3 x 1L/1000cm3
= 0.950 L
Initial concentration of NH3 = moles/Volume
= 0.180/0.950 = 0.1895 M
Initial concentration of NH4Cl = 0.04674/0.950
= 0.0492 M
The balanced reaction with ICE TABLE
NH4+ = NH3 + H+
I 0.0492 0.1895
C - x +x +x
E (0.0492-x) (0.1895+x) x
Equilibrium constant expression of the reaction
Ka = [H+] [NH3] / [NH4+]
5.714*10^-10 = (x) (0.1895+x) / (0.0492-x)
x << 0.0492
5.714*10^-10 = (x) (0.1895) / (0.0492)
x = 1.484 * 10^-10
[H+] = 1.484 * 10^-10
pH = - log [H+] = - log (1.484 * 10^-10)
= 9.83
Part c
Hypothetical species
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