   # A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 6, Problem 101AE
Textbook Problem
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## A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution.

Interpretation Introduction

Interpretation:

The concentration of the sodium hydroxide solution has to be calculated.

Concept Introduction:

The precipitate will be formed when two soluble salt solutions combined together is called as a precipitation reaction.

The only possible precipitate is Cr(OH)3

Cr(NO3)3(aq)+3NaOH(aq)Cr(OH)+3(s)3NaNO3(aq)

One mole of chromium (III) nitrate is reacts with 3 moles of NaOH to give 1 mole of Cr(OH)3 and 3 mole of NaNO3

Mole=volumeofsolution (L)×molarity (M)litre (L) (1)

### Explanation of Solution

Explanation

To determine: mole of NaOH used to form precipitate

2.06gCr(OH)3×1moleCr(OH)3103.02g×3moleNaOHmoleCr(OH)3

=6.00×10-2mole

To determine: mole of NaOH used to react with HCl

NaOH(aq)+HCl(aq)NaCl(aq)+H2O(l)

0.1000L×0

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