A mixture is prepared using 29.00 mL of a 0.0852 M weak base (pKa 4.580), 29.00
ID: 1058736 • Letter: A
Question
A mixture is prepared using 29.00 mL of a 0.0852 M weak base (pKa 4.580), 29.00 mL of a o.0639 M w Map acid (pKa 3.220) and 1.000 mL of O.000105 M HIn and then diluting to 100.0 mL, where HIn is the protonated indicator. The absorbance measured at 550 nm in a 5.000 cm cell was 0.1069. The molar absorptivity (e) values for HIn and its deprotonated form In- at 550 nm are 2.26 x 104 M-1cm-1 and 1.53 x 104 M-1cm 1 respectively. What is the pH of the solution? Number What are the concentrations of In and HIn? Number In M Number HIn M What is the pKa for Hln? Number pKaExplanation / Answer
base pKb = 14 - pKa
= 14 - 4.580
= 9.420
Kb = 10^-pKb
Kb = 3.80 x 10^-10
Ka = 10^-3.220
Ka = 6.03 x 10^-4
B + HA ---------------------> BH+ + A-
equalibrium constnat (K ) = Ka x Kb / Kw
=6.03 x 10^-4 x 3.80 x 10^-10 / 1.0 x 10^-14
= 22.9
molarity of base in dilute solution = 29 x 0.0852 / 100
= 0.0247 M
molarity of acid in dilute solution = 29 x 0.0639 / 100
= 0.0185 M
B + HA ---------------------> BH+ + A-
0.0247 0.0185 0 0 ----------------> initial
0.0247- x 0.0185-x x x ---------------> equilibrium
K = [BH+][A-]/[B][HA]
22.9 = x^2 / (0.0247-x ) ( 0.0185-x)
21.9 x^2 - 0.989 x + 0.01048 = 0
x = 0.0170
[HA] = 0.0185- x = 1.5 x 10^-3 M
[A-] = 0.0170 M
pH = pKa + log [A- / HA]
pH = 3.220 + log (0.0170 / 1.5x 10^-3)
pH = 4.27
[HIn] + [In-] = 0.000105 x 1 / 100 = 1.05 x 10^-6
[HIn] + [In-] = 1.05 x 10^-6 -----------------------> (1)
1.13 x 10^5 [HIn] + 7.65 x 10^4 [In-] = 0.1069 -------------------> (2)
[HIn] = 7.28 x 10^-7 M
[In-] = 3.22 x 10^-7 M
pH = pKa + log [In- / HIn]
4.27 = pKa + log (3.22 x 10^-7 / 7.28 x 10^-7)
pKa = 4.62
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