Equilibrium and Ka Calculations

Equilibrium and Ka Calculations

Assessment

Interactive Video

Chemistry

11th - 12th Grade

Hard

Created by

Jackson Turner

FREE Resource

The video tutorial explains how to calculate the acid dissociation constant (Ka) of nitrous acid (HNO2) using its pH value. It begins with an introduction to the problem, followed by setting up the acid-base equilibrium and writing the Ka expression. The tutorial then demonstrates the use of the ICE box method to determine equilibrium concentrations. Finally, it shows how to use the given pH to find the hydronium concentration and calculate the Ka value, emphasizing the stoichiometric relationships and arithmetic involved.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the initial concentration of the HNO2 solution used in the calculations?

0.0345 M

0.0516 M

0.0234 M

0.0678 M

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which species are involved in the equilibrium expression for the reaction of nitrous acid with water?

Nitrous acid and hydronium

Nitrous acid and water

Hydronium and nitrite ions

Water and nitrite ions

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is water not included in the Ka expression for the reaction?

Because it is a gas

Because it is a pure liquid

Because it is a solid

Because it is an aqueous species

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the stoichiometric ratio of nitrous acid to hydronium in the reaction?

2:2

1:1

2:1

1:2

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the change in concentration for the nitrous acid in the ICE box setup?

-X

0

-2X

+X

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the hydronium concentration at equilibrium determined from the pH?

By subtracting the pH from 10

By adding 10 to the pH

By taking 10 to the power of the negative pH

By multiplying the pH by 10

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the equilibrium concentration of hydronium ions if the pH is 2.34?

0.46 M

0.046 M

0.00046 M

0.0046 M

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