You use a pH meter to determine Ka for the new monoprotic weak acid that you dis

Question

You use a pH meter to determine Ka for the new monoprotic weak acid that you discovered. At 25 degree C the pH was 3.54 +/- 0.01. What is the uncertainty in the concentration of H+ in the solution? If the initial concentration acid was 1M what would be the Ka if the equilibrium concentration of H+ is given by exactly pH 3.54 Since the concentration of H+ does have uncertainty, what are the upper and lower values of Ka that could result from this uncertainty? What is the range for the pKa values?

Explanation / Answer

Here pH =3.54+/- 0.01 , pH may 3.55 or 3.54 or 3.53 now calculate the [H+] as follows:

[H+] = 10^-pH

[H+] = 10^-5.53 = 2.95x10^-6 M
[H+] = 10^-5.54 = 2.88x10^-6 M
[H+] = 10^-5.55 = 2.81x10^-6 M

here the uncertainty is in the tenths column.

The molarity of hydrogen ions is 2.8x10^-6M ± 1x10^-7M .

Given that HA = 1.0M

pH = 3.54, so [H+] = 10^-5.54 = 2.88x10^-6 M

Ka = [H+][A-]/[HA]

Ka= 2.88x10^-6 M *2.88x10^-6 M /1.0
K a= 8.3*10^-12

Low Ka with lower limit:

pH = 3.53, so [H+] = 10^-5.53 = 2.95x10^-6 M

Ka = [H+][A-]/[HA]

Ka= 2.95x10^-6 M *2.95x10^-6 M /1.0
K a= 8.70*10^-12

Low Ka with upper limit:

pH = 3.55, so [H+] = 10^-5.55 = 2.81x10^-6 M

Ka = [H+][A-]/[HA]

Ka= 2.81x10^-6 M *2.81x10^-6 M /1.0
K a= 7.90*10^-12; pKa = 11.10

K a= 8.3*10^-12; pKa = 11.08

K a= 8.70*10^-12; pKa = 11.06

Thus the range for values id 11.06 to 11.10.

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