Math  /  Algebra

QuestionHA is a weak acid. A buffer containing 0.25 M HA and 0.50 MA - has a pH of 6.12. What is the Ka of the acid? 1.52×1061.52 \times 10^{-6} 5.82 7.59×1077.59 \times 10^{-7} 6.12 3.79×1073.79 \times 10^{-7} 7.59×1057.59 \times 10^{-5} none of the above

Studdy Solution

STEP 1

What is this asking? We need to find the acid dissociation constant (KaK_a) of a weak acid (HA) given the concentrations of the acid and its conjugate base (A^-) in a buffer solution, along with the pH of the buffer. Watch out! Don't forget to convert the pH to [H+][H^+] before plugging it into the KaK_a expression.
Also, remember that a buffer uses a weak acid, not a strong acid!

STEP 2

1. Calculate the hydrogen ion concentration.
2. Set up the KaK_a expression.
3. Calculate the KaK_a value.

STEP 3

We know the pH of the buffer solution is **6.12**.
The pH is defined as pH=log10([H+])pH = -\log_{10}([H^+]), where [H+][H^+] is the hydrogen ion concentration.
To find [H+][H^+], we can rearrange this equation: [H+]=10pH[H^+] = 10^{-pH}.

STEP 4

Let's plug in our **pH value** of **6.12**: [H+]=106.12[H^+] = 10^{-6.12} [H+]7.59×107 M[H^+] \approx 7.59 \times 10^{-7} \text{ M}So, our hydrogen ion concentration is approximately **7.59×1077.59 \times 10^{-7} M**.

STEP 5

The dissociation of our weak acid, HA, can be represented as: HAH++AHA \rightleftharpoons H^+ + A^- The KaK_a expression, which represents the equilibrium constant for this dissociation, is given by: Ka=[H+][A][HA]K_a = \frac{[H^+] \cdot [A^-]}{[HA]} where [HA][HA] is the concentration of the undissociated acid, [A][A^-] is the concentration of the conjugate base, and [H+][H^+] is the hydrogen ion concentration.

STEP 6

We are given that the buffer contains **0.25 M** HA and **0.50 M** A^-.
We also just calculated that [H+]7.59×107[H^+] \approx 7.59 \times 10^{-7} M.
Let's plug these values into our KaK_a expression: Ka=(7.59×107)(0.50)0.25K_a = \frac{(7.59 \times 10^{-7}) \cdot (0.50)}{0.25} Ka=3.795×1070.25K_a = \frac{3.795 \times 10^{-7}}{0.25}Ka1.52×106K_a \approx 1.52 \times 10^{-6}

STEP 7

The KaK_a of the weak acid HA is approximately 1.52×1061.52 \times 10^{-6}.

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