Chapter 4, Problem 92QP

For each of the following pairs of combinations, indicate which one will produce the greater mass of solid product:

$\begin{array}{l}\begin{array}{l}\left(\text{a}\right)\hfill \\ \text{32}\text{.75 mL 1}\text{.005 }M{\text{ Hg}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{and 40}\text{.75 mL 0}\text{.9885 }M\text{ NaCl}\hfill \\ \text{or}\hfill \\ \text{21}\text{.45 mL 0}\text{.9995 }M{\text{ Hg}}_{\text{2}}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{and 41}\text{.00 mL 1}\text{.245 }M\text{ NaCl}\hfill \\ \left(\text{b}\right)\hfill \\ \text{45}\text{.25 mL 0}\text{.8895 }M\text{ Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\text{and 175}\text{.4 mL 0}\text{.2440 }M\text{ NaOH}\hfill \\ \text{or}\hfill \\ \text{51}\text{.50 mL 0}\text{.8545 }M\text{ Ca}{\left({\text{C2H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\text{and 225}\text{.8 mL 0}\text{.2211 }M\text{ NaOH}\hfill \\ \left(\text{c}\right)\hfill \\ \text{248}\text{.2 mL 1}\text{.095 }M{\text{ AgNO}}_{\text{3}}\text{and 25}\text{.00 mL 2}\text{.010 }M\text{ Ba}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\hfill \\ \text{or}\hfill \end{array}\\ \text{250}\text{.5 mL 0}\text{.4095 }M{\text{ AgNO}}_{\text{3}}\text{and 35}\text{.00 mL 1}\text{.475 }M\text{ Ba}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\end{array}$

Interpretation Introduction

Interpretation:

The combination producing the greater mass of a solid product is to be determined.

Concept Introduction:

In a precipitation reaction, two ionic compounds in solution react to form an insoluble compound called precipitate.

The formation of precipitates can be predicted by using solubility rules. The mass of the precipitate can be determined from the stoichiometry of the reaction.

The number of moles are calculated as follows:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

Answer to Problem 92QP

Solution:

(a) The highest mole amount is yielded by the second combination.

(b)The highest mole amount is yielded by the second combination.

(c)No formation of precipitates has taken place.

Explanation of Solution

a)

Combination $1$: $\text{32}\text{.75 mL 1}{\text{.005 M Hg}}_2{\left({\text{NO}}_3\right)}_2\text{and 40}\text{.75 mL 0}\text{.9885 M NaCl}$

Combination $2$: $\text{21}\text{.45 mL 0}{\text{.9995 M Hg}}_2{\left({\text{NO}}_3\right)}_2\text{and 41}\text{.00 mL 1}\text{.245 M NaCl}$

The net ionic equation taking place, in this case, is as follows:

${\text{Hg}}_2^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to {\text{Hg}}_2{\text{Cl}}_2\left(s\right)$

Combination 1:

From the balanced equation, one mole of ${\text{ Hg}}_2{\left({\text{NO}}_3\right)}_2$ reacts with two moles of $\text{NaCl}$.

First, calculate the moles of ${\text{ Hg}}_2{\left({\text{NO}}_3\right)}_2$ from the given volume and concentration as follows:

$\begin{array}{c}{\text{Moles of Hg}}_2{\left({\text{NO}}_3\right)}_2\text{=}\left(\frac{\text{ 1}{\text{.005 mol Hg}}_2{\left({\text{NO}}_3\right)}_2}{1\cancel{{\text{L Hg}}_2{\left({\text{NO}}_3\right)}_2}}\right)\times \text{0}\text{.03275 }\cancel{{\text{L Hg}}_2{\left({\text{NO}}_3\right)}_2}\\ =0.03291375{\text{ mol Hg}}_2{\left({\text{NO}}_3\right)}_2\end{array}$

Moles of ${\text{Hg}}_2^{2+}$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Hg}}_2^{2+}\text{ = }\left(0.03291\cancel{{\text{mol Hg}}_2{\left({\text{NO}}_3\right)}_2}\times \frac{{\text{2 mol Hg}}_2^{2+}}{1\cancel{{\text{mol Hg}}_2{\left({\text{NO}}_3\right)}_2}}\right)\\ =0.06582{\text{ mol Hg}}_2^{2+}\end{array}$

Moles of ${\text{Hg}}_2{\text{Cl}}_2$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Hg}}_2{\text{Cl}}_2\text{ = }\left(0.06582\text{ m}\cancel{{\text{ol Hg}}_2^{2+}}\times \frac{1{\text{ mol Hg}}_2{\text{Cl}}_2}{1\cancel{{\text{mol Hg}}_2^{2+}}}\right)\\ =0.06582{\text{ mol Hg}}_2{\text{Cl}}_2\end{array}$

Now, the moles of $\text{NaCl}$ are calculated as follows:

$\begin{array}{c}\text{Moles of NaCl = }\left(\frac{\text{0}\text{.9885 mol NaCl}}{1\cancel{\text{L}}\cancel{\text{ NaCl}}}\times \text{0}\text{.04075 }\cancel{\text{L}}\cancel{\text{NaCl}}\right)\\ =0.0402814\text{ mol NaCl}\end{array}$

Moles of ${\text{Cl}}^{-}$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Cl}}^{-}\text{ = }\left(0.04028\cancel{\text{mol NaCl}}\times \frac{1\text{}{\text{ mol Cl}}^{-}}{1\cancel{\text{mol NaCl}}}\right)\\ =0.04028{\text{ mol Cl}}^{-}\end{array}$

Moles of ${\text{Hg}}_2{\text{Cl}}_2$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Hg}}_2{\text{Cl}}_2\text{ = }\left(0.04028\text{ m}\cancel{{\text{ol Cl}}^{-}}\times \frac{1{\text{ mol Hg}}_2{\text{Cl}}_2}{\text{2 }\cancel{{\text{mol Cl}}^{-}}}\right)\\ =0.02014{\text{ mol Hg}}_2{\text{Cl}}_2\end{array}$

Combination 2:

First, calculate the moles of ${\text{ Hg}}_2{\left({\text{NO}}_3\right)}_2$ from the given volume and concentration as follows:

$\begin{array}{c}{\text{Moles of Hg}}_2{\left({\text{NO}}_3\right)}_2\text{ =}\left(\frac{\text{ 0}{\text{.9995 mol Hg}}_2{\left({\text{NO}}_3\right)}_2}{1\cancel{{\text{L Hg}}_2{\left({\text{NO}}_3\right)}_2}}\times \text{0}\text{.02145 }\cancel{{\text{L Hg}}_2{\left({\text{NO}}_3\right)}_2}\right)\\ =0.02144{\text{ mol Hg}}_2{\left({\text{NO}}_3\right)}_2\end{array}$

Moles of ${\text{Hg}}_2^{2+}$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Hg}}_2^{2+}\text{ = }\left(0.02144\cancel{{\text{mol Hg}}_2{\left({\text{NO}}_3\right)}_2}\times \frac{{\text{2 mol Hg}}_2^{2+}}{1\cancel{{\text{mol Hg}}_2{\left({\text{NO}}_3\right)}_2}}\right)\\ =0.04288{\text{ mol Hg}}_2^{2+}\end{array}$

Moles of ${\text{Hg}}_2{\text{Cl}}_2$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Hg}}_2{\text{Cl}}_2\text{ = }\left(0.04288\text{ m}\cancel{{\text{ol Hg}}_2^{2+}}\times \frac{1{\text{ mol Hg}}_2{\text{Cl}}_2}{1\cancel{{\text{mol Hg}}_2^{2+}}}\right)\\ =0.04288{\text{ mol Hg}}_2{\text{Cl}}_2\end{array}$

Now, the moles of $\text{NaCl}$ are calculated as follows:

$\begin{array}{c}\text{Moles of NaCl = }\left(\frac{\text{1}\text{.245 mol NaCl}}{1\cancel{\text{L}}\cancel{\text{NaCl}}}\times \text{0}\text{.04100 }\cancel{\text{L}}\cancel{\text{NaCl}}\right)\\ =0.05105\text{ mol NaCl}\end{array}$

Moles of ${\text{Cl}}^{-}$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Cl}}^{-}\text{ = }\left(0.05105\cancel{\text{mol NaCl}}\times \frac{1\text{}{\text{ mol Cl}}^{-}}{1\cancel{\text{mol NaCl}}}\right)\\ =0.05105{\text{ mol Cl}}^{-}\end{array}$

Moles of ${\text{Hg}}_2{\text{Cl}}_2$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Hg}}_2{\text{Cl}}_2\text{ = }0.05105\text{ m}\cancel{{\text{ol Cl}}^{-}}\times \frac{1{\text{ mol Hg}}_2{\text{Cl}}_2}{\text{2 }\cancel{{\text{mol Cl}}^{-}}}\\ =0.02552{\text{ mol Hg}}_2{\text{Cl}}_2\end{array}$

Therefore, the secondcombination will produce a greater mass of the product according to limiting reactant, and therefore, the highest mass of precipitates is yielded by this combination.

b)

Combination1: $\text{45}\text{.25 mL 0}\text{.8895 M Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\text{ and 175}\text{.4 mL 0}\text{.2440 M NaOH}$

Combination 2: $\text{51}\text{.50 mL 0}\text{.8545 M Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\text{ and 225}\text{.8 mL 0}\text{.2211 M NaOH}$:

Combination 1:

The precipitation reaction between ${\text{Ca}}_{\text{2}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)$ and $\text{NaOH}$ is:

$\text{Ca}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\left(aq\right)+2\text{NaOH}\left(aq\right)\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+{\text{2NaC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$

From the balanced equation, one mole of ${\text{Ca}}_{\text{2}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)$ reacts with two moles of NaOH.

First, calculate the moles of ${\text{Ca}}_{\text{2}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)$ from the given volume and concentration.

$\begin{array}{c}\text{Moles of Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\text{ =}\left(\frac{\text{ 0}\text{.8895 mol Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}{1\cancel{\text{L Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}}\times \text{0}\text{.04525 }\cancel{\text{L Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}\right)\\ =0.0402499\text{ mol Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\end{array}$

Moles of ${\text{Ca}}_{}^{2+}$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Ca}}_{}^{2+}\text{ = }\left(0.04025\cancel{\text{mol Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}\times \frac{{\text{1 mol Ca}}_{}^{2+}}{1\cancel{\text{mol Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}}\right)\\ =0.04025{\text{ mol Ca}}_{}^{2+}\end{array}$

Moles of $\text{Ca}\left({\text{OH}}_2\right)$ are calculated as follows:

$\begin{array}{c}\text{Moles of Ca}{\left(\text{OH}\right)}_2\text{ = }0.04025\text{ m}\cancel{{\text{ol Ca}}_{}^{2+}}\times \frac{1\text{ mol Ca}{\left(\text{OH}\right)}_2}{\text{1 }\cancel{{\text{mol Ca}}_{}^{2+}}}\\ =0.04025\text{ mol Ca}\left({\text{OH}}_2\right)\end{array}$

Now, calculate the moles of $\text{NaOH}$ from the given volume and concentration.

$\begin{array}{c}\text{Moles of NaOH =}\frac{\text{ 0}\text{.2440 mol NaOH}}{1\cancel{\text{L NaOH}}}\times \text{ 0}\text{.1754 }\cancel{\text{L NaOH}}\\ =0.0428\text{ mol NaOH}\end{array}$

Moles of ${\text{OH}}^{\text{-}}$ are calculated as follows:

$\begin{array}{c}{\text{Moles of OH}}^{-}\text{ = }0.0428\cancel{\text{mol NaOH}}\times \frac{1\text{}{\text{ mol OH}}^{-}}{1\cancel{\text{mol NaOH}}}\\ =0.0428{\text{ mol OH}}^{-}\end{array}$

Here, $\text{NaOH}$ is a limiting reagent that determines the amount of product formed.

$\begin{array}{c}\text{Moles of Ca}{\left(\text{OH}\right)}_{\text{2}}\text{ = 0}\text{.0428 }\cancel{{\text{mol OH}}^{-}}\times \frac{\text{1 mol Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{2 }\cancel{{\text{mol OH}}^{-}}}\\ =0.0214\text{ mol Ca}{\left(\text{OH}\right)}_{\text{2}}\end{array}$

Combination 2:

First, calculate the moles of ${\text{Ca}}_{\text{2}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)$ from the given volume and concentration.

$\begin{array}{c}\text{Moles of Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\text{ =}\frac{\text{ 0}\text{.8545 mol Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}{1\cancel{\text{L Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}}\times \text{ 0}\text{.0515o }\cancel{\text{L Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}\\ =0.044007\text{ mol Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\end{array}$

Moles of ${\text{Ca}}_{}^{2+}$ are calculated as follows:

$\begin{array}{c}{\text{Moles of Ca}}_{}^{2+}\text{ = }0.044007\cancel{\text{mol Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}\times \frac{{\text{1 mol Ca}}_{}^{2+}}{1\cancel{\text{mol Ca}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2}}\\ =0.044007{\text{ mol Ca}}_{}^{2+}\end{array}$

Moles of $\text{Ca}\left({\text{OH}}_2\right)$ are calculated as follows:

$\begin{array}{c}\text{Moles of Ca}{\left(\text{OH}\right)}_2\text{ = }0.044007\text{ m}\cancel{{\text{ol Ca}}_{}^{2+}}\times \frac{1\text{ mol Ca}{\left(\text{OH}\right)}_2}{\text{1 }\cancel{{\text{mol Ca}}_{}^{2+}}}\\ =0.044007\text{ mol Ca}\left({\text{OH}}_2\right)\end{array}$

Now, calculate the moles of $\text{NaOH}$ from the given volume and concentration.

$\begin{array}{c}\text{Moles of NaOH =}\frac{\text{ 0}\text{.2211 mol NaOH}}{1\cancel{\text{L NaOH}}}\times \text{ 0}\text{.2258 }\cancel{\text{L NaOH}}\\ =0.04992\text{ mol NaOH}\end{array}$

Moles of ${\text{OH}}^{\text{-}}$ are calculated as follows:

$\begin{array}{c}{\text{Moles of OH}}^{-}\text{ = }0.04992\cancel{\text{mol NaOH}}\times \frac{1\text{}{\text{ mol OH}}^{-}}{1\cancel{\text{mol NaOH}}}\\ =0.04992{\text{ mol OH}}^{-}\end{array}$

Here, $\text{NaOH}$ is a limiting reagent that determines the amount of product formed.

$\begin{array}{c}\text{Moles of Ca}{\left(\text{OH}\right)}_{\text{2}}\text{ = 0}\text{.04992 }\cancel{{\text{mol OH}}^{-}}\times \frac{\text{1 mol Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{2 }\cancel{{\text{mol OH}}^{-}}}\\ =0.02496\text{ mol Ca}{\left(\text{OH}\right)}_{\text{2}}\end{array}$

Therefore, the second combination will produce a greater mass of the product according to limiting reactant, and therefore, the highest mass of precipitates is yielded by this combination.

c)

Combination1: $\text{248}\text{.2 mL 1}{\text{.095 M AgNO}}_3\text{ and 25}\text{.00 mL 2}\text{.010 M Ba}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2$

Combination 2: $\text{250}\text{.5 mL 0}\text{.4095}M{\text{ AgNO}}_3\text{ and 35}\text{.00 mL 1}\text{.475}M\text{ Ba}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2$

The precipitation reaction between silver nitrate and barium acetate is:

$\text{Ba}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\left(aq\right)+2{\text{AgNO}}_3\left(aq\right)\to 2{\text{AgC}}_2{\text{H}}_3{\text{O}}_2\left(s\right)+\text{Ba}{\left({\text{NO}}_3\right)}_2\left(aq\right)$

In this reaction, barium acetate and silver nitrate reactwith each other. On the basis of solubility guidelines, acetates and nitrates are always soluble in each other. In this reaction, both will solubilize and no precipitation will take place as these both are soluble in each other.

Therefore, no precipitates will be formed.