# A 0.1236-g sample of an unknown monoprotic acid was dissolved in 25.8 mL of water and titrated...

## Question:

A 0.1236-g sample of an unknown monoprotic acid was dissolved in 25.8 mL of water and titrated with 0.0680 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 19.1 mL. After 11.7 mL of base had been added during the titration, the pH was determined to be 5.17. What is the Ka of the unknown acid?

## Henderson-Hasselbalch Equation:

The Henderson-Hasselbalch equation involves the value of the dissociation constant, the concentration of the species present in the buffer solution and the pH value of the buffer solution. So, any value can be calculated when the other parameters of the equation are given.

## Answer and Explanation: 1

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View this answerWe are given:

- Mass of acid (HA), {eq}\rm m_{HA} = 0.1236\ g {/eq}

- Volume of acid, {eq}\rm V_{HA} = 25.8\ mL {/eq}

- Molarity of {eq}\rm NaOH =...

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Chapter 24 / Lesson 4Learn what the Henderson-Hasselbalch equation is. See examples of the Henderson-Hasselbalch equation and understand how to utilize the equation with bases and acids.

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