Chapter 15, Problem 15.100QP

A 2.500-g sample of a mixture of sodium carbonate and sodium chloride is dissolved in 25.00 mL of 0.798 M HCl. Some acid remains after the treatment of the sample.

1. a Write the net ionic equation for the complete reaction of sodium carbonate with hydrochloric acid
2. b If 28.7 mL of 0.108 M NaOH were required to titrate the excess hydrochloric acid, how many moles of sodium carbonate were present in the original sample?
3. c What is the percent composition of the original sample?

Interpretation Introduction

Interpretation:

The explanations for the given set of statements have to be given.

Concept introduction:

The percent composition of the given original sample can be calculated by using the following formula.

${\text{percent composition = mol NaHCO}}_{\text{3}}\times \frac{{\text{g NaHCO}}_{\text{3}}}{{\text{1 mol NaHCO}}_{\text{3}}}$

Answer to Problem 15.100QP

The complete reaction is

${\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{ + 2H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + CO}}_{\text{2(g)}}$

Explanation of Solution

The complete reaction of sodium carbonate with hydrochloric acid is

${\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{ + 2H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + CO}}_{\text{2(g)}}$

Interpretation Introduction

Interpretation:

The explanations for the given set of statements have to be given.

Concept introduction:

The percent composition of the given original sample can be calculated by using the following formula.

${\text{percent composition = mol NaHCO}}_{\text{3}}\times \frac{{\text{g NaHCO}}_{\text{3}}}{{\text{1 mol NaHCO}}_{\text{3}}}$

Answer to Problem 15.100QP

The number of moles of sodium carbonate present in the original sample is $\text{0}\text{.00843 mol}.$

Explanation of Solution

First we can determine the total number of moles of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ from $\text{HCl}$ as follows.

${\text{Moles of H}}_{\text{3}}{\text{O}}^{\text{+}}\text{= }\frac{\text{0}\text{.798molHCl}}{\text{1L}}\text{× 0}\text{.02520L = 0}\text{.01995 mol}$

Now, let’s find the number of moles of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions that reacts with sodium hydroxide as follows.

${\text{Moles of H}}_{\text{3}}{\text{O}}^{\text{+}}\text{= }\frac{\text{0}\text{.109 mol NaOH}}{\text{1L}}\text{×0}\text{.0287L= 0}\text{.003100 mol}$

We come over that, the number of moles of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ present in the given original sample which is equal to the number of moles of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions which reacted with ${\text{CO}}_{\text{3}}{}^{\text{2-}}$ and is given as follows

${\text{Moles Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ = 1/2(total moles H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ - moles H}}_{\text{3}}{\text{O}}^{\text{+}}\text{ reacted with the NaOH)}$

$\text{= 1/2(0}\text{.01995 mol - 0}\text{.003100 mol) = 0}\text{.008425 = 0}\text{.00843 mol}$

Hence, the number of moles of sodium carbonate present in the original sample is $\text{0}\text{.00843 mol}$

Interpretation Introduction

Interpretation:

The explanations for the given set of statements have to be given.

Concept introduction:

The percent composition of the given original sample can be calculated by using the following formula.

${\text{percent composition = mol NaHCO}}_{\text{3}}\times \frac{{\text{g NaHCO}}_{\text{3}}}{{\text{1 mol NaHCO}}_{\text{3}}}$

Answer to Problem 15.100QP

The percent composition of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ original sample is $\text{35}\text{.7\%}\text{.}$

The percent composition of $\text{NaCl}$ original sample is $\text{64}\text{.3\%}.$

Explanation of Solution

To determine the percent composition, first determine the mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ present in original sample as follows.

$\text{0}{\text{.008425 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\times \frac{\text{105}{\text{.99g Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{{\text{1 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}}=\text{0}\text{.8929 g}$

Now, the percent ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the original sample that was given by

${\text{percent Na}}_{\text{2}}{\text{CO}}_{\text{3}}=\frac{\text{0}\text{.8929g}}{\text{2}\text{.500g}}\times 100\%=\text{35}\text{.71 = 35}\text{.7\%}$

Hence, the percent composition of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ original sample is $\text{35}\text{.7\%}.$

$\text{Percent NaCl = 100 - 35}\text{.71 = 64}\text{.29 = 64}\text{.3\%}$

Hence, the percent composition of $\text{NaCl}$ original sample is $\text{ 64}\text{.3\%}.$