Answered

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.

a. 0.700 mole of [tex]$Ca\left(NO_3\right)_2[/tex] in 800.0 mL of solution
\[
\begin{array}{l}
M_{Ca^{2+}} = \square \, M \\
M_{NO_3^{-}} = \square \, M
\end{array}
\]

b. 8.5 moles of [tex]$Na_2SO_4[/tex] in 4.50 L of solution
[tex]\[
\begin{array}{l}
M_{Na^{+}} = \square \, M \\
M_{SO_4^{2-}} = \square \, M
\end{array}
\][/tex]

c. 6.40 g of [tex]$NH_4Cl[/tex] in 760.0 mL of solution
\[
\begin{array}{l}
M_{NH_4^{+}} = \square \, M \\
M_{Cl^{-}} = \square \, M
\end{array}
\]

d. 2.60 g of [tex]$K_3PO_4[/tex] in 940.0 mL of solution
[tex]\[
\begin{array}{l}
M_{K^{+}} = \square \, M \\
M_{PO_4^{3-}} = \square \, M
\end{array}
\][/tex]



Answer :

GinnyAnswer
To address the problem, we will calculate the concentration of ions present in solutions of strong electrolytes step-by-step.

Part (a):

Given:
- 0.700 moles of [tex]\( \text{Ca(NO}_3\text{)}_2 \)[/tex] in 800.0 mL of solution.

First, let's find the concentration of [tex]\( \text{Ca(NO}_3\text{)}_2 \)[/tex] in moles per liter (M).
[tex]\[ M_{\text{Ca(NO}_3\text{)}_2} = \frac{0.700 \text{ moles}}{0.800 \text{ L}} = 0.875\text{ M} \][/tex]

1 mol of [tex]\( \text{Ca(NO}_3\text{)}_2 \)[/tex] dissociates into 1 mol of [tex]\( \text{Ca}^{2+} \)[/tex] and 2 mol of [tex]\( \text{NO}_3^- \)[/tex].

Thus:
[tex]\[ \text{M}_{\text{Ca}^{2+}} = 0.875\text{ M} \][/tex]
[tex]\[ \text{M}_{\text{NO}_3^-} = 2 \times 0.875\text{ M} = 1.75\text{ M} \][/tex]

Part (b):

Given:
- 8.5 moles of [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] in 4.50 L of solution.

First, let's find the concentration of [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] in moles per liter (M).
[tex]\[ M_{\text{Na}_2\text{SO}_4} = \frac{8.5 \text{ moles}}{4.50 \text{ L}} = 1.888\text{ M} \][/tex]

1 mol of [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] dissociates into 2 mol of [tex]\( \text{Na}^+ \)[/tex] and 1 mol of [tex]\( \text{SO}_4^{2-} \)[/tex].

Thus:
[tex]\[ \text{M}_{\text{Na}^+} = 2 \times 1.888\text{ M} = 3.778\text{ M} \][/tex]
[tex]\[ \text{M}_{\text{SO}_4^{2-}} = 1.888\text{ M} \][/tex]

Part (c):

Given:
- 6.40 g of [tex]\( \text{NH}_4\text{Cl} \)[/tex] in 760.0 mL of solution.
- Molar mass of [tex]\( \text{NH}_4\text{Cl} \)[/tex] = 53.49 g/mol.

First, let's find the number of moles of [tex]\( \text{NH}_4\text{Cl} \)[/tex].
[tex]\[ \text{mol}_{\text{NH}_4\text{Cl}} = \frac{6.40 \text{ g}}{53.49 \text{ g/mol}} = 0.1197 \text{ mol} \][/tex]

Next, let's find the concentration in moles per liter (M).
[tex]\[ M_{\text{NH}_4\text{Cl}} = \frac{0.1197 \text{ mol}}{0.760 \text{ L}} = 0.1574\text{ M} \][/tex]

1 mol of [tex]\( \text{NH}_4\text{Cl} \)[/tex] dissociates into 1 mol of [tex]\( \text{NH}_4^+ \)[/tex] and 1 mol of [tex]\( \text{Cl}^- \)[/tex].

Thus:
[tex]\[ \text{M}_{\text{NH}_4^+} = 0.1574\text{M} \][/tex]
[tex]\[ \text{M}_{\text{Cl}^-} = 0.1574\text{M} \][/tex]

Part (d):

Given:
- 2.60 g of [tex]\( \text{K}_3\text{PO}_4 \)[/tex] in 940.0 mL of solution.
- Molar mass of [tex]\( \text{K}_3\text{PO}_4 \)[/tex] = 212.27 g/mol.

First, let's find the number of moles of [tex]\( \text{K}_3\text{PO}_4 \)[/tex].
[tex]\[ \text{mol}_{\text{K}_3\text{PO}_4} = \frac{2.60 \text{ g}}{212.27 \text{ g/mol}} = 0.01225 \text{ mol} \][/tex]

Next, let's find the concentration in moles per liter (M).
[tex]\[ M_{\text{K}_3\text{PO}_4} = \frac{0.01225 \text{ mol}}{0.940 \text{ L}} = 0.01303\text{ M} \][/tex]

1 mol of [tex]\( \text{K}_3\text{PO}_4 \)[/tex] dissociates into 3 mol of [tex]\( \text{K}^+ \)[/tex] and 1 mol of [tex]\( \text{PO}_4^{3-} \)[/tex].

Thus:
[tex]\[ \text{M}_{\text{K}^+} = 3 \times 0.01303\text{ M} = 0.03909\text{ M} \][/tex]
[tex]\[ \text{M}_{\text{PO}_4^{3-}} = 0.01303\text{ M} \][/tex]

Here's the final answer:

Part (a):

[tex]\[ \text{M}_{\text{Ca}^{2+}} = 0.875\text{ M} \][/tex]
[tex]\[ \text{M}_{\text{NO}_3^-} = 1.75\text{ M} \][/tex]

Part (b):

[tex]\[ \text{M}_{\text{Na}^+} = 3.778\text{ M} \][/tex]
[tex]\[ \text{M}_{\text{SO}_4^{2-}} = 1.888\text{ M} \][/tex]

Part (c):

[tex]\[ \text{M}_{\text{NH}_4^+} = 0.1574\text{ M} \][/tex]
[tex]\[ \text{M}_{\text{Cl}^-} = 0.1574\text{ M} \][/tex]

Part (d):

[tex]\[ \text{M}_{\text{K}^+} = 0.03909\text{ M} \][/tex]
[tex]\[ \text{M}_{\text{PO}_4^{3-}} = 0.01303\text{ M} \][/tex]