Question

acid, the ratio of conjugate base to acid can be determined. At pH = pK a for an acid, [conjugate base] = [acid] in solution. At pH >> pK a for an acid, the acid will be mostly ionized. At pH << pK a for an acid, the acid will be mostly ionized. View Available Hint(s) Which statements regarding the Henderson-Hasselbalch equation are true? 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. At pH = pK a for an acid, [conjugate base] = [acid] in solution. At pH >> pK a for an acid, the acid will be mostly ionized. At pH << pK a for an acid, the acid will be mostly ionized. All of the listed statements are true. 1, 2, and 3 are true. 2, 3, and 4 are true. 1, 2, and 4 are true.

Answer 1

Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

• 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. TRUE

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If you know pH and pka:

10^(pH-pka) = $\frac{[A^-]}{[HA]}$

The ratio will be: 10^(pH-pka)

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

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

0 = log₁₀ $\frac{[A^-]}{[HA]}$

10^0 = $\frac{[A^-]}{[HA]}$

1 = $\frac{[A^-]}{[HA]}$

As ratio is 1,  [conjugate base] = [acid] in solution.

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

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

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

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!