Question

What are the concentrations of hydroxide and hydronium ions in a solution with a pH of 10.2? 1.4 × 10–4 M H3O+ and 7.1 × 10–11 M OH– 3.8 × 10–6 M H3O+ and 2.6 × 10–9 M OH– 8.3 × 10–9 M H3O+ and 1.2 × 10–6 M OH– 6.3 × 10–11 M H3O+ and 1.6 × 10–4 M OH–

Answer 1

Answering:

It is D

Explanation:

Just took the test

Answer 2

The concentration of hydronium ion [H₃O⁺] and hydroxide ion [OH¯] are:

1. Concentration of hydronium ion [H₃O⁺] is 6.3×10¯¹¹ M

2. Concentration of Hydroxide ion [OH¯] is 1.6×10¯⁴ M

The pH of a solution is simply a measure of the acidic or alkalinity of the solution.

The concentration of hydronium ion [H₃O⁺] and hydroxide ion [OH¯] can be obtained as illustrated below:

Step 1

Data obtained from the question

pH = 10.2

Concentration of hydronium ion [H₃O⁺] =…?

Concentration of Hydroxide ion [OH¯] =…?

Step 2

Determination of the concentration of hydronium ion [H₃O⁺]

pH = 10.2

Concentration of hydronium ion [H₃O⁺] =…?

pH = –log[H₃O⁺]

10.2 =  –log[H₃O⁺]

Multiply though by –1

–10.2 = log[H₃O⁺]

Take antilog of –10.2

[H₃O⁺] = antilog(–10.2)

[H₃O⁺] = 6.3×10¯¹¹ M

Step 3:

Determination of the pOH

pH = 10.2

pOH =?

pH + pOH = 14

10.2 + pOH = 14

Collect like terms

pOH = 14 – 10.2

pOH = 3.8

Step 4

Determination of the concentration of Hydroxide ion [OH¯].

pOH = 3.8

Concentration of Hydroxide ion [OH¯] =…?

pOH = –log[OH¯]

3.8 = –log[OH¯]

Multiply through by –1

–3.8 = log[OH¯]

Take the antilog of –3.8

[OH¯] = antilog(–3.8)

[OH¯] = 1.6×10¯⁴ M

SUMMARY:

1. Concentration of hydronium ion [H₃O⁺] = 6.3×10¯¹¹ M

2. Concentration of Hydroxide ion [OH¯] = 1.6×10¯⁴ M

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