Which statments regarding the henderson-hasselbalch equation are true?

Complete question is;

Which statements regarding the Henderson-Hasselbalch equation are true?

1. If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined.

2. At pH = pKa for an acid, [conjugate base] = [acid] in solution.

3. At pH > pKa for an acid, the acid will be mostly ionized.

4. At pH < pKa for an acid, the acid will be mostly ionized.

A. All of the listed statements are true. B. 1, 2, and 3 are true.

C. 2, 3, and 4 are true.

D. 1, 2, and 4 are true.

B. 1, 2, and 3 are true.

Explanation:

The formula for the Henderson-Hasselbalch equation is:

pH = pka + log₁₀([A^(-)]/[HA])

Where;

PH is acidity of solution

ka is acid dissociation constant

A^(-) is concentration of conjugate base

HA is concentration of Acid

- For statement 1; If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined;

pH = pka + log₁₀([A^(-)]/[HA])

pH - pka = log₁₀([A^(-)]/[HA])

10^(pH - pka) = ([A^(-)]/[HA])

Since we can find the ratio as seen, then the statement is true

- For statement 2: At pH = pKa for an acid, [conjugate base] = [acid] in solution;

We will substitute pH for pKa;

pH = pH + log₁₀([A^(-)]/[HA])

This give;

0 = log₁₀([A^(-)]/[HA])

10^(0) = [A^(-)]/[HA]

1 = [A^(-)]/[HA]

Thus; [A^(-)] = [HA]

Thus, the statement is true

- For statement 3: At pH > pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is greater than 0 and thus;

10^(pH - pKa) is greater than 1.

Thus;

[A^(-)]/[HA] > 1

[A^(-)] > [HA]

So more acid is ionized than base.

So the statement is true.

- For statement 4: At pH < pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is less than 0 and thus;

10^(pH - pKa) is less than 1.

Thus;

[A^(-)]/[HA] < 1

[A^(-)] < [HA]

So we have more base ionized than acid. So statement is false

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