What is the concentration of H⁺ in a solution with pH = 2.50?

Question

Quizplus · Accepted Answer

The concentration of HS1U1P1+S1S1P0 can be calculated using the Henderson-Hasselbalch equation: 
pH = pKa + log([A-]/[HA])
where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

In this case, HS1U1P1+S1S1P0 is the acid and its conjugate base is S1S1P0-. The pKa of this acid can be found in a reference book or online. Let's assume it is 7.5.

At pH 2.5, the concentration of [A-]/[HA] can be calculated as follows: 
pH = pKa + log([A-]/[HA])
2.5 = 7.5 + log([S1S1P0-]/[HS1U1P1+S1S1P0])
-5 = log([S1S1P0-]/[HS1U1P1+S1S1P0])
10^-5 = [S1S1P0-]/[HS1U1P1+S1S1P0]

We know that the total concentration of HS1U1P1+S1S1P0 in the solution is equal to [HS1U1P1+S1S1P0] + [S1S1P0-]. Let's call this concentration "x".

We can use the equation above to substitute for [S1S1P0-]: 
10^-5 = [x - HS1U1P1+S1S1P0]/[HS1U1P1+S1S1P0]

Solving for [HS1U1P1+S1S1P0], we get: 
[HS1U1P1+S1S1P0] = x / (1 + 10^-5)

We know that the pH is very acidic (pH = 2.5), which means the concentration of HS1U1P1+S1S1P0 is likely to be high. Therefore, we can assume that the concentration of [S1S1P0-] is negligible compared to [HS1U1P1+S1S1P0]. This assumption is valid if the ratio of [S1S1P0-]/[HS1U1P1+S1S1P0] is less than 0.1.

In this case, we can check whether this assumption is valid: 
[S1S1P0-]/[HS1U1P1+S1S1P0] = 10^-5
This is less than 0.1, so our assumption is valid.

Therefore, we can simplify our expression for [HS1U1P1+S1S1P0]: 
[HS1U1P1+S1S1P0] ≈ x

We are looking for the concentration of HS1U1P1+S1S1P0, which is approximately equal to the total concentration of acid-base species in the solution: 
[HS1U1P1+S1S1P0] ≈ x = [HS1U1P1+S1S1P0] + [S1S1P0-]

Solving for x, we get: 
x = 10^-pH = 10^-2.5 = 0.0032 M

Therefore, the concentration of HS1U1P1+S1S1P0 in the solution is approximately 0.0032 M, which corresponds to choice C.

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