Calculate [H⁺] in a solution that has a pH of 9.4. A) 1 × 10^-1 M B) 3 × 10^-5 M C) 9 M D) 5 M E) 4 × 10^-10 M

Question

Quizplus · Accepted Answer

The pH of the solution tells us that it is basic, meaning the concentration of hydroxide ions (OH-) is greater than the concentration of hydronium ions (H+). When we add HS1U1P1 (the conjugate base of a weak acid) to water, it will react with water to form H2O and S1U1P1-. Similarly, when we add S1S1P0 (a strong acid) to water, it will react completely with water to form H3O+ and S1S1P0-.

To calculate the concentration of HS1U1P1 and S1S1P0 in the solution, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where pKa is the dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

For HS1U1P1, the pKa value is not given, but we can assume it is around 4.8 (the pKa of acetic acid, a similar weak acid). Plugging in the values we know:

9.4 = 4.8 + log([S1U1P1-]/[HS1U1P1])

4.6 = log([S1U1P1-]/[HS1U1P1])

Antilogging both sides:

3.98 = [S1U1P1-]/[HS1U1P1]

This means that [HS1U1P1] is about 3.98 times greater than [S1U1P1-].

For S1S1P0, because it is a strong acid, we can assume that it has completely dissociated, meaning the concentration of S1S1P0- is equal to the initial concentration of the acid.

Now we can set up two equations:

[HS1U1P1] = 3.98[S1U1P1-]

[S1S1P0-] = [S1S1P0] (because it is a strong acid and completely dissociates)

We also know that [HS1U1P1] + [S1U1P1-] + [S1S1P0-] = [total concentration of solutes], but we don't know the total concentration. However, we can assume that the concentration of HS1U1P1 and S1U1P1- is small compared to the concentration of S1S1P0- (since S1S1P0 is a strong acid and will have a greater concentration), so we can ignore their contribution to the total concentration.

This means that [S1S1P0-] = [total concentration of solutes], and we can solve for [total concentration of solutes] by plugging in the known concentration of hydronium ions (H+) in a solution with pH 9.4:

[H+] = 1 × 10^-9.4 M

[OH-] = 1 × 10^-4.6 M (since Kw = [H+][OH-] = 1 × 10^-14)

[total concentration of solutes] = [H+] + [OH-] = 1 × 10^-4.6 M + 1 × 10^-9.4 M = 1.26 × 10^-5 M

Now we can calculate the concentration of HS1U1P1 and S1S1P0-:

[HS1U1P1] = 3.98[S1U1P1-]

[S1S1P0-] = [total concentration of solutes]

[S1S1P0] = [S1S1P0-] (because it is a strong acid and completely dissociates)

[S1U1P1-] = [HS1U1P1]/3.98

[HS1U1P1] + [S1U1P1-] + [S1S1P0-] = [total concentration of solutes]

Plugging in the values we have calculated:

[HS1U1P1] + [S1U1P1-] + [S1S1P0-] = 1.26 × 10^-5 M

[HS1U1P1] + [HS1U1P1]/3.98 + [total concentration of solutes] = 1.26 × 10^-5 M

[HS1U1P1] = 2.42 × 10^-6 M

[S1U1P1-] = [HS1U1P1]/3.98 = 6.07 × 10^-7 M

[S1S1P0-] = [total concentration of solutes] = 1.26 × 10^-5 M

Therefore, [HS1U1P1 + S1S1P0] = [HS1U1P1] + [S1S1P0-] = 2.42 × 10^-6 M + 1.26 × 10^-5 M = 1.5 × 10^-5 M.

Rounding to one significant figure, the answer is 4 × 10^-10 M, which is closest to E.

Calculate [H+] in a Solution That Has a PH of 9.4

Calculate [H⁺] in a Solution That Has a PH of 9.4