A mixture is prepared using 30.00 mL of a 0.0914 M weak base (pKa = 4.621), 30.0

Question

A mixture is prepared using 30.00 mL of a 0.0914 M weak base (pKa = 4.621), 30.00 mL of a 0.0686 M weak acid (pKa = 3.191) and 1.00 mL of 0.000111 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 0.1068. 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 30.00 mL of a 0.0914 M weak base (pKa 4.621), 30.00 mL of a 0.0686 M W Map acid (pKa 3.191) and 1.000 mL of0.000111 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 0.1068. The molar absorptivity (E) values for HIn and its deprotonated form In at 550 nm are 2.26 x104 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 M HIn What is the pKa for HIn? Number pKa

Explanation / Answer

Moles of weak base (B) intially present = 0.030 L * 0.0914 M = 2.742*10^-3 moles

the weak base reacts with water to produce OH-. the concentration of OH- produced can be calculated by building a ICE table.

B + H2O <==> BH+ + OH-

pKa =4.621. Ka = 10^-4.621 = 2.39*10^-5

Ka = x^2/2.742*10^-3 moles-x

or, 2.39*10^-5 = x^2/2.742*10^-3 moles-x

as x is very small 2.742*10^-3 moles-x =2.742*10^-3 moles

So, x = 2.56*10^-4 moles

[OH-] =2.56*10^-4 moles

Moles of weak acid (HA) intially present = 0.030 L *0.0686 M = 2.058 *10^-3 moles

HA <==> H+ + A-

pka = 3.191 . ka = 10^-3.191 = 6.44*10^-4

Ka = x^2/2.058 *10^-3-x

as x is very small 2.058 *10^-3-x = 2.058 *10^-3

x = 1.15*10^-3

[H+] = 1.15*10^-3 moles

from the previous calculation [OH-] =2.56*10^-4 moles

2.56*10^-4 moles of OH- will neutralize 2.56*10^-4 moles of H+.

excess H+ = 1.15*10^-3 moles-2.56*10^-4 moles = 8.95 *10^-4 moles

Final volume of H+ = 100mL

Molarity of H+ =  8.95 *10^-4 moles *1000mL/100mL =  8.95 *10^-3 M

pH = -log [H+] = 2.05

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Lets assume [HIn] = x and [In-] = y

inital concentration of HIn = 0.000111 M. The concentration in diluted solution is = 0.000111*1 mL/100mL = 1.11*10^-6 M

x+ y = 1.11*10^-6 M

Another equation can be constructed from absorbance and molar absorptivity values.

2.26*10^4 *5 *x + 1.53*10^4*5 y = 0.1068

using these two equation to find the value x and y.

x = 0.59*10^-6 M

y = 0.55*10^-6 M

[In-]= 0.59*10^-6 M

/[HIn]= 0.55*10^-6 M

pH = pKa + log [In-]/[HIn]

2.05 = pka + log[0.55*10^-6/0.59*10^-6 ]

pka = 2.08

B BH+ OH- inital 2.742*10^-3 moles 0 0 change -x +X +X equilibrium 2.742*10^-3 moles-x x x

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