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# Calculate [OH ] given [H3O+] in each aqueous solution and ISBN: 9780321910295 34

## Solution for problem 63P Chapter 14

Introductory Chemistry | 5th Edition

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Problem 63P

Problem 63P

Calculate [OH ] given [H3O+] in each aqueous solution and classify the solution as acidic or basic.

(a) [H3O+] = 1.5 × 10−9 M

(b) [H3O+] = 9.3 × 10−9 M

(c) [H3O+] = 2.2 × 10−6 M

(d) [H3O+] = 7.4 × 10−4 M

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate hydroxyl ion concentration in each of the solution.

Step1:

In pure water, one H2O molecule donates proton and acts as an acid and another water molecule accepts a proton and acts as a base at the same time. The following equilibrium exists:

H2O(l) + H2O(l) ------> H3O+(aq) + OH-(aq)

The equilibrium constant for the reaction is given by:

Kw= [H3O+][OH]        -----(1)

Where, Kw is called the ionic product of water and its value is 1 x 10-14 M2.

We can distinguish acidic, neutral and basic aqueous solutions by the relative values of the H3O+ and OH concentrations:

Acidic: [H3O+] > [OH]

Neutral: [H3O+] = [OH]

Basic: [H3O+] < [OH]

Step2:

1. Given, [H3O+] = 1.5 x 10-9 M

From equation (1), [OH] = Kw / [H3O+]

[OH] = (1 x 10-14 M2) / (1.5 x 10-9 M)

[OH] = 6.67 x 10-6 M

Since, [H3O+] < [OH], therefore, the solution is basic.

b)        Given, [H3O+] = 9.3 x 10-9 M

From equation (1), [OH] = Kw / [H3O+]

[OH] = (1 x 10-14 M2) / (9.3 x 10-9 M)

[OH] = 1.07 x 10-6 M

Since, [H3O+] < [OH], therefore, the solution is basic.

c)        Given, [H3O+] = 2.2 x 10-6 M

From equation (1), [OH] = Kw / [H3O+]

[OH] = (1 x 10-14 M2) / (2.2 x 10-6 M)

[OH] = 4.54 x 10-9 M

Since, [H3O+] > [OH], therefore, the solution is acidic.

d)        Given, [H3O+] = 7.4 x 10-4 M

From equation (1), [OH] = Kw / [H3O+]

[OH] = (1 x 10-14 M2) / (7.4 x 10-4 M)

[OH] = 1.35 x 10-11 M

Since, [H3O+] > [OH], therefore, the solution is acidic.

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Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

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Calculate [OH ] given [H3O+] in each aqueous solution and