A mixture is prepared using 27.00 mL of a 0.0817 M weak base (pKa = 4.573), 27.0
ID: 1059042 • Letter: A
Question
A mixture is prepared using 27.00 mL of a 0.0817 M weak base (pKa = 4.573), 27.00 mL of a 0.0613 M weak acid (pKa = 3.190) and 1.00 mL of 0.000101 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 0.1015. 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.
What is the pH of the solution? Number pH What are the concentrations of In and HIn? Number Number M What is the pKa for Hln? Number pKaExplanation / Answer
base pKb = 14 - pKa
= 14 - 4.573
= 9.427
Kb = 10^-pKb
Kb = 3.71 x 10^-10
Ka = 10^-3.190
Ka = 6.46 x 10^-4
B + HA ---------------------> BH+ + A-
equalibrium constnat (K ) = Ka x Kb / Kw
= 6.46 x 10^-4 x 3.71 x 10^-10 / 1.0 x 10^-14
= 24.0
molarity of base in dilute solution = 27 x 0.0817 / 100
= 0.0221 M
molarity of acid in dilute solution = 27 x 0.0613 / 100
= 0.0166 M
B + HA ---------------------> BH+ + A-
0.0221 0.0166 0 0 ----------------> initial
0.0221-x 0.0166-x x x ---------------> equilibrium
K = [BH+][A-]/[B][HA]
24.0 = x^2 / (0.0221-x ) ( 0.0166-x)
23 x^2 - 0.9288 x + 8.8 x 10^-3 = 0
x = 0.0252 , 0.0152
x = 0.0152
[HA] = 0.0166 - 0.0152 = 1.4 x 10^-3 M
[A-] = 0.0152 M
pH = pKa + log [A- / HA]
pH = 3.190 + log (0.0152 / 1.4x 10^-3)
pH = 4.23
[HIn] + [In-] = 0.000101 x 1 / 100 = 1.01 x 10^-6
[HIn] + [In-] = 1.01 x 10^-6 -----------------------> (1)
1.13 x 10^5 [HIn] + 7.65 x 10^4 [In-] = 0.1015 -------------------> (2)
[HIn] = 6.64 x 10^-7 M
[In-] = 3.46 x 10^-7 M
pH = pKa + log [In- / HIn]
4.23 = pKa + log (3.46 x 10^-7 / 6.64 x 10^-7)
pKa = 4.51
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