A mixture is prepared using 30.00 mL of a 0.0849 M weak base (pKa = 4.556), 30.0

Question

A mixture is prepared using 30.00 mL of a 0.0849 M weak base (pKa = 4.556), 30.00 mL of a 0.0637 M weak acid (pKa = 3.250) and 1.00 mL of 0.000119 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 0.1140. The molar absorptivity (?) values for HIn and its deprotonated form In at 550 nm are 2.26 104 M1cm1 and 1.53 104 M1cm1, respectively.

1) what is pH of the solution

2) what are the concentrations of In- and HIn

3) what is the pKa for HIn

Explanation / Answer

1)

given

30 ml of a 0.0637 M weak acid

it is diluted to 100 ml

so

using

M1V1 =M2V2

0.0637 x 30 = y x 100

y = 0.01911 M

so

the new conc of acid is 0.01911 M

now for a weak acid

[H+] =sqrt (Ka x C)

also

pKa = -log Ka

so

3.25 =-log Ka

Ka = 5.62 x 10-4

so

[H+] = sqrt ( 5.62 x 10-4 x 0.01911)

[H+]= 3.28 x 10-3 M

similarly for the weak base

after dilution

M1V1 = M2V2

0.0849 x 30 = y x 100

y = 0.02547 M

now for weak base

[OH-] = sqrt ( Kb x C)

given

pKa = 4.556

-log Ka = 4.556

Ka = 2.78 x 10-5

now

Kb = 10-14 / Ka

Kb = 10-14 / 2.78 x 10-5

Kb = 3.60 x 10-10

so

[OH-] = sqrt (3.60 x 10-10 x 0.02547)

[OH] = 3.027 x 10-6

now

[H+] = 10-14 / [OH-]

so

[H+] = 10-14 / 3.027 x 10-6

[H+] = 3.30 x 10-9

the [H+] contribution from HIn cane neglected

so

total [H+] = 3.28 x 10-3 + 3.3 x 10-9

[H+] = 3.28 x 10-3

pH = -log [H+]

pH = -log 3.28 x 10-3

pH = 2.484

so the pH is 2.484

2)

given 1 ml of 0.000119 M HIn

after dilution

M1V1 = M2V2

0.000119 x 1 = Y x 100

Y = 1.19 x 10-6 M

so

the new conc of HIn is 1.19 x 10-6

now

HIn ---> H+ + In-

let HIn reacted = y

then

[In-] formed = y

[HIn] remaining = 1.19 x 10-6 - y

we know that

according to beers law

A = e1 x l x c1 + e2 x l x c2

given

path length (l) = 5

e1= 2.26 x 10^4

c1 = 1.19 x 10-6 - y

e2 = 1.53 x 10^4

c2 = y

so

0.114 = [ 2.6 x 10^4 x 5 x ( 1.19 x 10-6 - y) ] + ( 1.53 x 10^4 x 5 x y )

0.114 = 0.1547 - ( 1.3 x 10^5) y + ( 7.65 x 10^4) y

solving we get

y = 7.607 x 10-7

so

the concentrations are

[In-] = 7.607 x 10-7

[HIn] = 1.19 x 10-6 - 7.607 x 10-7 = 4.29 x 10-7

3)

now

HIn ---> H+ + In-

from the above we can see that

[H+] = [In-] = 7.607 x 10-7

now

ka = [H+] [In-] / [HIn]

Ka = 7.607 x 10-7 x 7.607 x 10-7 / 4.29 x 10-7

Ka =1.348 x 10-6

so

pKa = -logKa

pKa = -log 1.348 x 10-6

pKa = 5.87

so

the pKa for HIn is 5.87

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