IDNLearn.com: Your trusted platform for finding reliable answers. Our Q&A platform offers detailed and trustworthy answers to ensure you have the information you need.
Sagot :
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]
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]
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Thank you for visiting IDNLearn.com. We’re here to provide accurate and reliable answers, so visit us again soon.
