Question

To study a marine organism, a biologist prepares a 1.00-kg sample to simulate the ion concentrations in seawater. She mixes 26.5 g of NaCl, 2.40 g of MgCl2, 3.35 g of MgSO4, 1.20 g of CaCl2, 1.05 g of KCl, 0.315 g of NaHCO3, and 0.098 g of NaBr in distilled water.

(a) If the density of the solution is 1.025 g/cm3, what is the molarity of each ion?

(b) What is the total molarity of alkali metal ions?

(c) What is the total molarity of alkaline earth metal ions?

(d) What is the total molarity of anions?

Short answer

(a)the molarity of each ion is

$$\begin{gathered} {\text{InNaCl,Na}}^{\text{+}}{\text{=0.465,Cl}}^{-}\text{=0.465} \\ \text{In Mg}{\text{Cl}}_{\text{2,}}{\text{Mg}}^{\text{2 +}}{\text{=2.58 x10}}^{\text{- 2}}{\mathrm{Cl}}^{-}={\text{5.16×10}}^{\text{- 2}} \\ {\text{InMgSO}}_{\text{4}}{\text{,Mg}}^{\text{2 +}}{\text{=2.85×10}}^{\text{-2}}{{\text{SO}}_{\text{4}}}^{\text{2-}}{\text{=2.85×10}}^{\text{-2}} \\ {\text{In CaCl}}_{\text{2,}}{\text{,Ca}}^{\text{2+}}{\text{=1.11×10}}^{\text{- 2}}{\text{Cl}}^{\text{-}}{\text{=2.22×10}}^{\text{-2}} \\ {\text{In KCl,K}}^{\text{+}}{\text{=1.44×10}}^{\text{-2}}{\text{Cl}}^{\text{-}}{\text{=1.44×10}}^{\text{-2}} \\ {\text{In NaHCO}}_{\text{3}}{\text{Na}}^{\text{+}}{\text{=3.84×10}}^{\text{-3}}\text{HC}{{\text{O}}_{\text{3}}}^{\text{-}}{\text{=3.84×10}}^{\text{-3}} \\ \mathrm{In}{\text{NaBrNa}}^{\text{+}}{\text{=9.76×10}}^{\text{-4}}{\text{Br}}^{\text{-}}{\text{=9.76×10}}^{\text{-4}} \end{gathered}$$

(b)the total molarity of alkali metal ions is$0.484M$

(c)the total molarity of alkaline earth metal ions is$6.54x{10}^{-2}M$

(d)the total molarity of anions is$0.587M$

Step-by-step solution

the given values are:

26.5 g of NaCl

2.40 g of MgCl₂

3.35 g of MgSO₄

1.20 g of CaCl₂

1.05 g of KCl

0.315 g of NaHCO₃

0.098 g of NaBr

Now do the calculation part, first, find out the volume and then molarity of the solution and we can easily find out the other values also as follows: 

$\begin{array}{rcl}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathit{v}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathit{o}\mathit{f}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{a}\mathit{m}\mathit{p}\mathit{l}\mathit{e}& \mathbf{=}& \left(1000gm\right)\mathbf{\left(}\frac{\mathbf{1}\mathbf{c}{\mathbf{m}}^{\mathbf{3}}}{\mathbf{1}\mathbf{.}\mathbf{025}\mathbf{g}\mathbf{m}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{l}\mathbf{i}\mathbf{t}}{\mathbf{1000}\mathbf{c}{\mathbf{m}}^{\mathbf{3}}}\mathbf{\right)}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{976}\mathit{l}\mathit{i}\mathit{t}\\ & & \end{array}$

Molarity can find out by dividing moles by volume:

For NaCl

$\begin{array}{rcl}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{C}\mathit{l}& \mathbf{=}& \frac{\left(26.5gmNaCl\right)\mathbf{\left(}\frac{\mathbf{1}\mathbf{N}\mathbf{a}\mathbf{C}\mathbf{l}}{\mathbf{58}\mathbf{.}\mathbf{44}\mathbf{g}\mathbf{m}\mathbf{N}\mathbf{a}\mathbf{C}\mathbf{l}}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{976}\mathbf{l}\mathbf{i}\mathbf{t}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{465}\mathit{x}{\mathbf{10}}^{\mathbf{22}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{l}\mathit{i}\mathit{t}\\ & & \end{array}$

The molarity of each ion is:

$$\begin{gathered} \mathit{N}{\mathit{a}}^{\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(0.465\frac{molNaCl}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}{\mathbf{a}}^{\mathbf{+}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{C}\mathbf{l}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{465}\mathit{M}\mathit{N}{\mathit{a}}^{\mathbf{+}} \\ \mathit{C}{\mathit{l}}^{\mathbf{-}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(0.465\frac{molNaCl}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{C}{\mathbf{l}}^{\mathbf{-}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{C}\mathbf{l}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{465}\mathit{M}\mathit{C}{\mathit{l}}^{\mathbf{-}} \end{gathered}$$

For MgCl₂

$\begin{array}{rcl}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{o}\mathit{f}\mathit{M}\mathit{g}\mathit{C}{\mathit{l}}_{\mathbf{2}}& \mathbf{=}& \frac{\left(2.40gmMgC{l}_2\right)\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}}{\mathbf{95}\mathbf{.}\mathbf{21}\mathbf{g}\mathbf{m}\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{976}\mathbf{l}\mathbf{i}\mathbf{t}}\\ & \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{58}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{l}\mathit{i}\mathit{t}\\ & & \end{array}$

The molarity of each ion is:

$$\begin{gathered} \mathit{M}{\mathit{g}}^{\mathbf{2}\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(2.58x{10}^{-2}\frac{molMgC{l}_2}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{M}{\mathbf{g}}^{\mathbf{2}\mathbf{+}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{58}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}\mathit{M}{\mathit{g}}^{\mathbf{2}\mathbf{+}} \\ \mathit{C}{\mathit{l}}^{\mathbf{-}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(2.58x{10}^{-2}\frac{molMgC{l}_2}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{2}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{C}{\mathbf{l}}^{\mathbf{-}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{16}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}\mathit{C}{\mathit{l}}^{\mathbf{-}} \end{gathered}$$

The molarity of each ion is:

For MgSO₄

$\begin{array}{rcl}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{o}\mathit{f}\mathit{M}\mathit{g}\mathit{S}{\mathit{O}}_{\mathbf{4}}& \mathbf{=}& \frac{\left(3.35gmMgS{O}_4\right)\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{M}\mathbf{g}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}}{\mathbf{120}\mathbf{.}\mathbf{38}\mathbf{g}\mathbf{m}\mathbf{M}\mathbf{g}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{976}\mathbf{l}\mathbf{i}\mathbf{t}}\\ & \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{85}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{l}\mathit{i}\mathit{t}\\ & & \end{array}$

The molarity of each ion is:

$$\begin{gathered} \mathit{M}{\mathit{g}}^{\mathbf{2}\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(2.85x{10}^{-2}\frac{molMgS{O}_4}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{M}{\mathbf{g}}^{\mathbf{2}\mathbf{+}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{M}\mathbf{g}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{85}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}\mathit{M}{\mathit{g}}^{\mathbf{2}\mathbf{+}} \\ \mathit{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{-}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(2.85x{10}^{-2}\frac{molMgS{O}_4}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{2}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{C}{\mathbf{l}}^{\mathbf{-}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{M}\mathbf{g}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{85}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}\mathit{S}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{-}} \end{gathered}$$

For CaCl₂

$\begin{array}{rcl}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{o}\mathit{f}\mathit{C}\mathit{a}\mathit{C}{\mathit{l}}_{\mathbf{2}}& \mathbf{=}& \frac{\left(1.20gmCaC{l}_2\right)\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{C}\mathbf{a}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}}{\mathbf{110}\mathbf{.}\mathbf{98}\mathbf{g}\mathbf{m}\mathbf{C}\mathbf{a}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{976}\mathbf{l}\mathbf{i}\mathbf{t}}\\ & & \\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{11}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{l}\mathit{i}\mathit{t}\\ & & \end{array}$

The molarity of each ion is:

$$\begin{gathered} \mathit{C}{\mathit{a}}^{\mathbf{2}\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(1.11x{10}^{-2}\frac{molCaC{l}_2}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{C}{\mathbf{a}}^{\mathbf{2}\mathbf{+}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{C}\mathbf{a}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{11}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}\mathit{C}{\mathit{a}}^{\mathbf{2}\mathbf{+}} \\ \mathit{C}{{\mathbf{l}}_{\mathbf{2}}}^{\mathbf{-}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(1.11x{10}^{-2}\frac{molCaC{l}_2}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{2}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{C}{\mathbf{l}}^{\mathbf{-}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{C}\mathbf{a}\mathbf{C}{\mathbf{l}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{22}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}\mathit{C}{\mathit{l}}^{\mathbf{-}} \end{gathered}$$

For KCl

$\begin{array}{rcl}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{o}\mathit{f}\mathit{K}\mathit{C}\mathit{l}& \mathbf{=}& \frac{\left(1.05gmKCl\right)\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{K}\mathbf{C}\mathbf{l}}{\mathbf{74}\mathbf{.}\mathbf{55}\mathbf{g}\mathbf{m}\mathbf{K}\mathbf{C}\mathbf{l}}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{976}\mathbf{l}\mathbf{i}\mathbf{t}}\\ & & \\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{44}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{l}\mathit{i}\mathit{t}\\ & & \end{array}$

The molarity of each ion is:

$$\begin{gathered} {\mathit{K}}^{\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(1.44x{10}^{-2}\frac{molKCl}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}{\mathbf{K}}^{\mathbf{+}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{K}\mathbf{C}\mathbf{l}}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{44}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}{\mathit{K}}^{\mathbf{+}} \\ \mathit{C}{\mathit{l}}^{\mathbf{-}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(1.44x{10}^{-2}\frac{molKCl}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{C}{\mathbf{l}}^{\mathbf{-}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{K}\mathbf{C}\mathbf{l}}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{44}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}\mathit{C}{\mathit{l}}^{\mathbf{-}} \end{gathered}$$

For NaHCO₃

_{$\begin{array}{rcl}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}& \mathbf{=}& \frac{\left(0.315gmNaHC{O}_3\right)\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}{\mathbf{84}\mathbf{.}\mathbf{01}\mathbf{g}\mathbf{m}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{976}\mathbf{l}\mathbf{i}\mathbf{t}}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{84}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{l}\mathit{i}\mathit{t}\\ & & \end{array}$}

The molarity of each ion is:

$$\begin{gathered} \mathit{N}{\mathit{a}}^{\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(3.84x{10}^{-3}\frac{molNaHC{O}_3}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}{\mathbf{a}}^{\mathbf{+}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{84}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{M}\mathit{N}{\mathit{a}}^{\mathbf{+}} \\ \mathit{H}\mathit{C}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(3.84x{10}^{-3}\frac{molNaHC{O}_3}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{H}\mathbf{C}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{H}\mathbf{C}{\mathbf{O}}_{\mathbf{3}}}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{84}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{M}\mathit{H}\mathit{C}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}} \end{gathered}$$

For NaBr

$\begin{array}{rcl}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathit{o}\mathit{f}\mathit{N}\mathit{a}\mathit{B}\mathit{r}& \mathbf{=}& \frac{\left(0.098gmNaBr\right)\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{B}\mathbf{r}}{\mathbf{102}\mathbf{.}\mathbf{89}\mathbf{g}\mathbf{m}\mathbf{N}\mathbf{a}\mathbf{B}\mathbf{r}}\mathbf{\right)}}{\mathbf{0}\mathbf{.}\mathbf{976}\mathbf{l}\mathbf{i}\mathbf{t}}\\ & & \\ & \mathbf{=}& \mathbf{9}\mathbf{.}\mathbf{76}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{l}\mathit{i}\mathit{t}\\ & & \end{array}$

The molarity of each ion is:

$$\begin{gathered} \mathit{N}{\mathit{a}}^{\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(9.76x{10}^{-4}\frac{molNaBr}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}{\mathbf{a}}^{\mathbf{+}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{B}\mathbf{r}}\mathbf{\right)}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{76}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M}\mathit{N}{\mathit{a}}^{\mathbf{+}} \\ \mathit{B}{\mathit{r}}^{\mathbf{-}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\left(9.76x{10}^{-4}\frac{molNaBr}{lit}\right)\mathit{x}\mathbf{\left(}\frac{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{B}{\mathbf{r}}^{\mathbf{-}}}{\mathbf{1}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{N}\mathbf{a}\mathbf{B}\mathbf{r}}\mathbf{\right)}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{76}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M}\mathit{B}{\mathit{r}}^{\mathbf{-}} \end{gathered}$$

(b) the alkali metals in the given mixture are Na and K, therefore the total molarity of alkali metal ions is:

$0.456M+1.44x{10}^{-2}M+3.84x{10}^{-3}M+9.76x{10}^{-4}M=0.484M$

(c) the alkali earth metals in the given mixture are Mg and Ca, therefore the total molarity of alkali metal ions is:

$2.58x{10}^{-2}M+2.85x{10}^{-2}M+1.11x{10}^{-2}M=6.54x{10}^{-2}M$

(d)the total molarity of anions in the mixture is:

$$\begin{gathered} 0.456M+5.16x{10}^{-2}M+2.85x{10}^{-2}M+2.22x{10}^{-2}+1.44x{10}^{-2}M \\ +3.84x{10}^{-3}M+9.76x{10}^{-4}M=0.587M \end{gathered}$$