Chapter 3, Problem 119AP

Calculate the number of cations and anions in each of the following compounds: (a) 8.38 g of $\text{KBr},\text{ (b) 5}{\text{.40 of Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{, (c) 7}{\text{.45 g of Ca}}_{\text{3}}{({\text{PO}}_4)}_2.$

Interpretation Introduction

Interpretation:

The number of cations and anions in the given compoundsis to be determined.

Concept introduction:

Equal number of anions and cations is present in an ionic compound to make the compound electrically neutral.

Positively charged ion is termed as cation whereas anion is negatively charged ion.

The number of moles of a reactant is calculated by the following formula:

$m=\frac{wt}{M_w}$

Here, $m$ is the number of moles of a compound, $wt$ is the weight of the compound, and $M_w$ is the molecular weight of the compound.

According to mole concept, one mole of substance contains $6.022\times {10}^{23}\text{ atoms}$.

Answer to Problem 119AP

Solution:

a)

$\text{0}\text{.424}\times {\text{10}}^{23}{\text{ K}}^{+}\text{ ions and 0}\text{.424}\times {\text{10}}^{23}{\text{ Br}}^{-}\text{ ions}$

b)

$\text{0}\text{.458}\times {\text{10}}^{23}{\text{ Na}}^{+}\text{ ions and 0}\text{.229}\times {\text{10}}^{23}{\text{ SO}}_4{}^{2-}\text{ ions}$

c)

$\text{0}\text{.434}\times {\text{10}}^{23}{\text{ Ca}}^{2+}\text{ ions and 0}\text{.289}\times {\text{10}}^{23}{\text{ (PO}}_4{)}^3{}^{-}\text{ ions}$

Explanation of Solution

a) $8.38\text{ g KBr}$

The number of moles of ions in $8.38\text{ g KBr}$ with the molecular weight of $119\text{ g}/\text{mole}$ is calculated as follows:

$\begin{array}{c}m=\left(\frac{8.38}{119}\right)\\ =0.070\text{ moles}\end{array}$

Thus, $0.070\text{ mole KBr}$ will have $0.070{\text{ mole K}}^{+}$ and $0.070{\text{ mole Br}}^{-}$ ions.

Now, in $1{\text{ mole of K}}^{+}\text{ ions}$, there are $6.022\times {10}^{23}{\text{ K}}^{+}\text{ ions}$.

The number of ${\text{K}}^{+}$ ions in $0.070{\text{ mole K}}^{+}$ is calculated as follows:

$\begin{array}{c}{\text{K}}^{+}\text{ ions}=\left(0.070\times 6.022\times {10}^{23}\right)\\ =0.424\times {10}^{23}\text{ ions}\end{array}$

The number of ${\text{Br}}^{-}$ ions in $0.070{\text{ mole Br}}^{-}$ is calculated as follows:

$\begin{array}{c}{\text{Br}}^{-}\text{ ions}=\left(0.070\times 6.022\times {10}^{23}\right)\\ =0.424\times {10}^{23}\text{ ions}\end{array}$

b) $\text{5}{\text{.40 g Na}}_2{\text{SO}}_4$

The number of moles of ions in $\text{5}{\text{.40 g Na}}_2{\text{SO}}_4$ with the molecular weight of $142\text{ g}/\text{mole}$ is calculated as follows:

$\begin{array}{c}m=\left(\frac{5.4}{142}\right)\\ =0.038\text{ moles}\end{array}$

Thus, $0.038{\text{ moles of Na}}_2{\text{SO}}_4$ will have $2\times 0.038{\text{ mole Na}}^{+}$ and $0.038{\text{ mole SO}}_4{}^{-}$ ions.

Now, in $1{\text{ mole Na}}^{+}$, there are $6.022\times {10}^{23}{\text{ Na}}^{+}\text{ ions}$.

The number of ${\text{Na}}^{+}$ ions in ${\text{Na}}_2{\text{SO}}_4$ is calculated as follows:

$\begin{array}{c}{\text{Na}}^{+}\text{ ions}=\left(2\times 0.038\times 6.022\times {10}^{23}\right)\\ =0.458\times {10}^{23}\text{ ions}\end{array}$

The number of ${\text{SO}}_4{}^{2-}$ ions in $0.038{\text{ mole SO}}_4{}^{2-}$ is calculated as follows:

$\begin{array}{c}{\text{SO}}_4{}^{2-}\text{ ions}=0.038\times 6.022\times {10}^{23}\\ =0.229\times {10}^{23}\text{ ions}\end{array}$

c) $7.45{\text{ g Ca}}_3{{\text{(PO}}_4\text{)}}_2$

The number of moles of ions in $7.45{\text{ g Ca}}_3{{\text{(PO}}_4\text{)}}_2$ with the molecular weight of $\text{310 g}/\text{mole}$ is calculated as follows:

$\begin{array}{c}m=\left(\frac{7.45}{310}\right)\\ =0.024\text{ moles}\end{array}$

Now, in ${\text{1 mole Ca}}^{2+}$, there are $6.022\times {10}^{23}{\text{ Ca}}^{2+}\text{ ions}$.

The number of ions in $\text{0}{\text{.024 mole Ca}}^{2+}$ is calculated as follows:

$\begin{array}{c}{\text{Ca}}^{+}\text{ ions}=\left(3\times 0.024\times 6.022\times {10}^{23}\right)\\ =0.434\times {10}^{23}\text{ ions}\end{array}$

The number of ions in $\text{0}{\text{.024 mole PO}}_4{}^{3-}$ is as follows:

$\begin{array}{c}{\text{PO}}_4{}^{3-}\text{ ions}=\left(2\times 0.024\times 6.022\times {10}^{23}\right)\\ =0.289\times {10}^{23}\text{ ions}\end{array}$