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Acid-Base Equilibrium Calculations

Acid-Base Equilibrium Calculations

Assessment

Interactive Video

Chemistry, Science, Mathematics

11th - 12th Grade

Practice Problem

Hard

Created by

Patricia Brown

FREE Resource

The video tutorial explains how to calculate the acid dissociation constant (KA) and the base dissociation constant (KB) using pH and pOH information. It provides a detailed walkthrough of calculating KA for hydrofluoric acid (HF) and KB for ethylamine, using given concentrations and equilibrium expressions. The tutorial emphasizes the importance of understanding equilibrium and dissociation in chemical solutions.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the primary purpose of using a pH meter in acid-base calculations?

To measure the temperature of the solution

To find the density of the solution

To determine the concentration of hydronium ions

To calculate the molarity of the solution

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the example of HF, what is the given concentration of hydronium ions?

0.116 M

0.00116 M

0.2 M

0.0116 M

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the equilibrium expression for the dissociation of HF in water?

[H3O+][HF]/[F-]

[H3O+][F-]/[HF]

[HF][H2O]/[H3O+]

[F-][H2O]/[HF]

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the calculated Ka value for HF in the example?

7.2 x 10^-3

7.2 x 10^-4

7.2 x 10^-5

7.2 x 10^-6

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What information is needed to calculate Kb from pOH?

Concentration of hydronium ions

Concentration of hydroxide ions

Concentration of water

Concentration of ethylamine

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the pOH value given for the ethylamine solution?

2.40

0.24

2.04

4.02

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you find the concentration of hydroxide ions from pOH?

By using the formula [OH-] = 10^-pOH

By using the formula [OH-] = pOH x 10

By using the formula [OH-] = pOH/10

By using the formula [OH-] = 10^pOH

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