A mixture is prepared using 26.00 mL of a 0.0917 M weak base (pKa = 4.591), 26.0
ID: 1058910 • Letter: A
Question
A mixture is prepared using 26.00 mL of a 0.0917 M weak base (pKa = 4.591), 26.00 mL of a 0.0688 M weak acid (pKa = 3.187) and 1.00 mL of 0.000116 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 (?) values for HIn and its deprotonated form In– at 550 nm are 2.26 × 104 M–1cm–1 and 1.53 × 104 M–1cm–1, respectively.
A mixture is prepared using 26.00 mL of a 0.091 M weak base (pKa 91), 26.00 mL of a 0.0688 M W Map DBb acid (pKa 3.187) and 1.00 mL of 0.000116 M HIn and then diluting to 100.0 mL, where HIn is the protonat 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 pH What are the concentrations of In and HIn? Number In M Number HIn M What is the pKa for HIn? Number pKaExplanation / Answer
base pKb = 14 - pKa
= 14 - 4.591
= 9.409
Kb = 10^-pKb
Kb = 3.90 x 10^-10
Ka = 10^-3.187
Ka = 6.50 x 10^-4
B + HA ---------------------> BH+ + A-
equalibrium constnat (K ) = Ka x Kb / Kw
= 6.50 x 10^-4 x 3.90 x 10^-10 / 1.0 x 10^-14
= 25.36
molarity of base in dilute solution = 26 x 0.0917 / 100
= 0.0238 M
molarity of acid in dilute solution = 26 x 0.0688 / 100
= 0.0179 M
B + HA ---------------------> BH+ + A-
0.0238 0.0179 0 0 ----------------> initial
0.0238-x 0.0179-x x x ---------------> equilibrium
K = [BH+][A-]/[B][HA]
25.36 = x^2 / (0.0238-x ) ( 0.0179-x)
24.36 x^2 - 1.0575 x + 0.0108 = 0
x = 0.0164
[HA] = 0.0179- x = 1.5 x 10^-3 M
[A-] = 0.0164 M
pH = pKa + log [A- / HA]
pH = 3.187 + log (0.0164 / 1.5x 10^-3)
pH = 4.23
[HIn] + [In-] = 0.00011y x 1 / 100 = 1.16 x 10^-6
[HIn] + [In-] = 1.16 x 10^-6 -----------------------> (1)
1.13 x 10^5 [HIn] + 7.65 x 10^4 [In-] = 0.1069 -------------------> (2)
[HIn] = 4.98 x 10^-7 M
[In-] = 6.62 x 10^-7 M
pH = pKa + log [In- / HIn]
4.23 = pKa + log (6.62 x 10^-7 / 4.98 x 10^-7)
pKa = 4.11
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