A mixture is prepared using 27.00 mL of a 0.0774 M weak base (pKa = 4.635), 27.0

Question

A mixture is prepared using 27.00 mL of a 0.0774 M weak base (pKa = 4.635), 27.00 mL of a 0.0581 M weak acid (pKa = 3.189) 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 0.1123. 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. 1.) What is the pH of the solution? 2.)What are the concentrations of In– and HIn? respectively 3.)What is the pKa for HIn?

Explanation / Answer

base   pKb = 14 - pKa

                  = 14 - 4.635

                   = 9.365

Kb = 10^-pKb

Kb = 4.32 x 10^-10

Ka = 10^-3.189

Ka = 6.47 x 10^-4

B   + HA ---------------------> BH+   + A-

equalibrium constnat (K ) = Ka x Kb / Kw

                                         = 6.47 x 10^-4 x 4.32 x 10^-10 / 1.0 x 10^-14

                                        = 27.96

molarity of base in dilute solution = 27.00 x 0.0774 / 100

                                                    = 0.0209 M

molarity of acid in dilute solution = 27.00 x 0.0581 / 100

                                                   = 0.0157 M

B        +    HA ---------------------> BH+   + A-

0.0209 0.0157 0         0 ----------------> initial

0.0209 -x   0.0157-x                       x          x ---------------> equilibrium

K = [BH+][A-]/[B][HA]

27.96 = x^2 / (0.0209 -x ) ( 0.0157-x)

27.96 x^2 + 9.17 x 10^-3 - 1.023 x = x^2

26.96 x^2 - 1.023 x + 9.17 x 10^-3 = 0

x = 0.0145

[HA] = 0.0157-x = 1.2 x 10^-3 M

[A-] = 0.0145 M

pH = pKa + log [A- / HA]

pH = 3.189 + log (0.0145 / 1.2x 10^-3)

pH = 4.27 -------------------------------------------------------------------------------> answer

[HIn] + [In-] = 0.000116 x 1 / 100 = 1.16 x 10^-6

[HIn] + [In-] = 1.16 x 10^-6 -----------------------> (1)

2.26 × 10^4 [HIn] + 1.53 × 10^4 [In-] = 0.1123 -------------------> (2)

[HIn] = 1.30 x 10^-5 M --------------------> answer

[In-] = 1.18 x 10^-5 M--------------------> answer

pH = pKa + log [In- / HIn]

4.27 = pKa + log (1.18 x 10^-5 / 1.30 x 10^-5)

pKa = 4.31 -----------------------------> answer

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