Chapter 1, Problem 1.38P

(a)

Interpretation Introduction

Interpretation:

Formula mass of acetic acid and sodium carbonate has to be calculated.

Concept Introduction:

Formula mass of a compound is equivalent to sum of atomic mass of the elements in the empirical formula of the compound.

Answer to Problem 1.38P

Formula mass of acetic acid is calculated to be $60.05g/mol.$

Formula mass of sodium bicarbonate is calculated to be $84.01g/mol.$

Explanation of Solution

Formula mass of acetic acid, $C{H}_{3}COOH$ is calculated as –

$\begin{array}{l}atomic\text{ mass of C}\text{= 12}\text{.0107 amu}\\ \text{atomic mass of H}\text{= 1}\text{.0008 amu}\\ \text{atomic mass of O}\text{= 15}\text{.999 amu}\end{array}$

In acetic acid $C{H}_{3}COOH$ molecule there are two C-atoms, four H-atoms and two O-atoms are present.  Sum of atomic mass of all Carbon, Hydrogen and  Oxygen atoms gives formula mass of acetic acid.  Hence formula mass of $C{H}_{3}COOH$ is,

$\begin{array}{l}\left(2\times atomic\text{ mass of C}\right)\text{ + }\left(4\times \text{atomic mass of H}\right)\text{ + 2(atomic mass of O)}\\ =\left(2\times 12.0107\text{ amu}\right)\text{ + }\left(\text{4}\times \text{1}\text{.008 amu}\right)\text{ + 2}\left(\text{15}\text{.999 amu}\right)\\ \text{= 60}\text{.05 amu = 60}\text{.05 g/mol}\end{array}$

In sodium bicarbonate $NaHC{O}_3$ molecule there are one C-atom, one H-atom, one Na-atom and three O-atoms are present. Sum of atomic mass of one Na-atom, one C-atom, one H-atom and three Oxygen atoms gives formula mass of sodium bicarbonate $NaHC{O}_3.$ Hence formula mass of $NaHC{O}_3$ is,

$\begin{array}{l}\left(1\times atomic\text{ mass of Na}\right)\text{ + }\left(1\times \text{atomic mass of H}\right)\text{ +1(atomic mass of C)}+\text{ 3(atomic mass of O)}\\ =\left(1\times 23\text{ amu}\right)\text{ + }\left(1\times \text{1}\text{.008 amu}\right)\text{ +1(12}\text{.0107 amu) + 3}\left(\text{15}\text{.999 amu}\right)\\ \text{= 60}\text{.05 amu = 84}\text{.01 g/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Amount of acetic acid to be reacted with $5\text{ g}$ of sodium bicarbonate has to be calculated.

Answer to Problem 1.38P

Amount of acetic acid to be reacted with $5\text{ g}$ of sodium bicarbonate is calculated to be $3.57g.$

Explanation of Solution

$\begin{array}{l}mol{\text{ NaHCO}}_3=\frac{mass{\text{ of NaHCO}}_3}{molar{\text{ mass of NaHCO}}_3}\\ =\frac{5\text{ g}}{84.01\text{ g/mol}}=0.0595\text{ mol}\end{array}$

The complete and balanced reaction between acetic acid and sodium bicarbonate is,

$C{H}_{3}COOH+NaHC{O}_3\stackrel{}{\to }C{H}_{3}COONa+C{O}_2+H_{2}O$

From the above equation we could observe one mole equivalent of bicarbonate react with one mole equivalent of acetic acid.  Hence $0.0595\text{ mol}$ of bicarbonate is converted into product by $0.0595\text{ mol}$ of acetic acid.  Thus number of grams of acetic acid that reacts with $5g$ of bicarbonate is,

$\begin{array}{l}number\text{ of grams}{\text{= mol of CH}}_3\text{COOH}\times {\text{molar mass of CH}}_3\text{COOH}\\ \text{= 0}\text{.0595 mol}\times \text{60}\text{.05 g/mol}\\ \text{= 3}\text{.57 g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Number of grams of $5\text{ wt\%}$ acetic acid to be reacted with $5\text{ g}$ of sodium bicarbonate has to be calculated.  Volume of $5\text{ wt\%}$ acetic acid to be reacted with $5\text{ g}$ of sodium bicarbonate has to be calculated.

Concept Introduction:

Weight percent is one of parameters used to express concentration of a solution.  It is expressed as,

$\text{Weight percent of a solute = }\frac{\text{weight of solute}}{\text{weight of solution}}\times \text{ 100\%}$

Answer to Problem 1.38P

Number of grams of $5\text{ wt\%}$ acetic acid to be reacted with $5\text{ g}$ of sodium bicarbonate is calculated as $71.4g\text{.}$

Volume of $5\text{ wt\%}$ acetic acid to be reacted with $5\text{ g}$ of sodium bicarbonate is calculated as $71.4mL\text{.}$

Explanation of Solution

$5\text{ wt\%}$ acetic acid contains $5\text{ mL}$ of acetic acid dissolved in $95\text{ mL}$ of distilled water.  Since density of acetic acid is $1.0\text{ g/mL,}$ mass of acetic acid in $5\text{ mL}$ is,

$\begin{array}{l}mass=\text{volume}\times \text{density}\\ \text{= 5 mL}\times \text{1}\text{.0 g/mL = 5 g}\end{array}$

Thus $100mL$ of $5\text{ wt\%}$ acetic acid solution contains $5g$ of acetic acid.  As calculated in the previous step $3.57\text{ g}$ of acetic acid reacts with $5g$ of bicarbonate.  Since we are using $5\text{ wt\%}$ acetic acid solution, number of grams of $5\text{ wt\%}$ acetic acid solution required to react with $5g$ of bicarbonate is,

$\frac{3.57\text{ g}}{0.05}=71.4\text{ g}$

Converting grams to milliliters,

$\begin{array}{l}volume=\frac{mass}{density}\\ \text{= }\frac{71.4\text{ g}}{1.0\text{ g/mL}}=71.4\text{ mL}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The limiting reagent has to be identified when $50mL$ of vinegar is reacted with $5g$ of sodium bicarbonate.

Concept Introduction:

In a chemical reaction, Limiting reagent is any one of the reactants which gets consumed completely while forming the product.  The number of moles of each reactant has to be calculated and compared to determine which one is the limiting reagent.

Answer to Problem 1.38P

Vinegar is the limiting reagent when $50mL$ of vinegar is reacted with $5g$ of sodium bicarbonate.

Explanation of Solution

$\begin{array}{l}volume\text{ of vinegar}=\text{ 50 mL}\\ \text{Mass of vinegar}\text{= 50 g, since density is 1}\text{.0 g/mL}\end{array}$

In previous steps we calculated $71.4\text{ mL}$ of vinegar is required to completely react with $5g$ of sodium bicarbonate.  But when $50\text{ mL}$ of vinegar is used, all of them are readily consumed by bicarbonate that none of them is left in the reaction.  Thus obviously  vinegar is the limiting reagent $50mL$ of vinegar is reacted with $5g$ of sodium bicarbonate.

(e)

Interpretation Introduction

Interpretation:

Quantity of $C{O}_2$ is generated in the reaction between acetic acid and sodium bicarbonate has to be calculated in liters given the conditions are $P=1\text{ bar, T=300K}\text{.}$  the pressure generated to pop the cork out of the bottle has to calculated.

Concept Introduction:

Ideal gas law relates Pressure, temperature, volume and number of moles of an ideal gas as,

$PV=nRT$

Where,

$\begin{array}{l}P=\text{ Pressure}\\ \text{V}\text{= Volume}\\ \text{T}\text{= Temperature}\\ \text{R}\text{= Universal gas constant}\\ \text{n}\text{= no}\text{.of moles}\end{array}$

Answer to Problem 1.38P

Quantity of $C{O}_2$ is generated for the given reaction is $0.005L.$

The pressure generated to pop the cork out of the bottle is calculated as $3bar.$

Explanation of Solution

One mole equivalent of $C{O}_2$ produced in the reaction.  Hence is number of moles is equivalent to that of sodium bicarbonate which is calculated as $0.0595\text{ mol}$ in the previous step.

Given data:

$\begin{array}{l}T=\text{ 300K}\\ \text{P}=\text{ 1 bar}\end{array}$

Volume of $C{O}_2$ generated is calculated as,

$\begin{array}{l}PV=\text{ nRT}\\ \text{V}\text{= n}\times \frac{RT}{P}\\ =\text{ 0}\text{.0595 mol}\times \frac{0.0831{\text{ L bar K}}^{-1}{\text{ mol}}^{-1}\times 300K}{1\text{ bar}}\\ =\text{ 0}\text{.005 L}\end{array}$

The bottle has $0.5L$ of air space and so the actual volume is $0.5L-0.005L=0.495L$ and the pressure generated to pop out the bottle is,

$\begin{array}{l}P=\text{ n}\times \frac{RT}{V}\\ =\text{ 0}\text{.0595 mol}\times \frac{0.0831{\text{ L bar K}}^{-1}{\text{ mol}}^{-1}\times 300K}{\text{0}\text{.495L }}\\ =\text{ 3 bar}\end{array}$