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Green and White H…
Kinetics and Equilibrium
Calculating the solubility of an ionic compound when a common ion is…

Calculate the solubility at of in pure water and in a solution. You’ll find data in the ALEKS Data tab.
Round both of your answers to 2 significant digits.
solubility in pure water:
solubility in solution:
(1]

Ask by Bryant Chandler. in the United States

Mar 26,2025

Question

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Green and White H…
Kinetics and Equilibrium
Calculating the solubility of an ionic compound when a common ion is…

Calculate the solubility at of in pure water and in a solution. You’ll find data in the ALEKS Data tab.
Round both of your answers to 2 significant digits.
solubility in pure water:
solubility in solution:
(1]

Ask by Bryant Chandler. in the United States

Mar 26,2025

UpStudy AI · Accepted Answer

**Step 1. Write the dissolution equilibrium and expression for $$K_{sp}$$.**

For $$\mathrm{BaCrO}_{4}$$ solid:
$$
\mathrm{BaCrO}_{4}(s) \rightleftharpoons \mathrm{Ba}^{2+}(aq) + \mathrm{CrO}_{4}^{2-}(aq)
$$
Thus, if the solubility (molar concentration of dissolved solid) is $$s$$ then at equilibrium:
$$
\mathrm{[Ba^{2+}]} = s \quad \text{and} \quad \mathrm{[CrO_{4}^{2-}]} = s.
$$
The solubility product expression is:
$$
K_{sp}= s \cdot s = s^{2}.
$$

We are given (from the ALEKS Data tab) that:
$$
K_{sp}= 1.17\times10^{-10}.
$$

---

**Step 2. Solubility in pure water**

For pure water, we have:
$$
s^{2}=1.17\times10^{-10}.
$$
Taking the square root:
$$
s=\sqrt{1.17\times10^{-10}}\approx 1.08\times10^{-5}~\mathrm{M}.
$$

The molar mass of $$\mathrm{BaCrO}_{4}$$ is calculated as:
$$
M_{\mathrm{BaCrO}_{4}} = M_{\mathrm{Ba}} + M_{\mathrm{Cr}} + 4 M_{\mathrm{O}} \approx 137.33 + 52.00 + 4(16.00) \approx 253.33~\mathrm{g/mol}.
$$

Now, convert the molar solubility to grams per liter:
$$
\text{Solubility (g/L)} = s \times M_{\mathrm{BaCrO}_{4}}\approx (1.08\times10^{-5}~\mathrm{mol/L})(253.33~\mathrm{g/mol})\approx 2.74\times10^{-3}~\mathrm{g/L}.
$$

Rounded to 2 significant digits:
$$
2.7\times10^{-3}~\mathrm{g/L}.
$$

---

**Step 3. Solubility in a $$0.0200~\mathrm{M}$$ $$\mathrm{BaCl}_{2}$$ solution**

In a solution of $$\mathrm{BaCl}_{2}$$, which dissociates completely,
$$
\mathrm{BaCl}_{2}\rightarrow \mathrm{Ba}^{2+}+ 2\mathrm{Cl}^{-},
$$
there is an initial $$\mathrm{Ba}^{2+}$$ concentration of $$0.0200~\mathrm{M}$$. Let the solubility of $$\mathrm{BaCrO}_{4}$$ in this solution be $$s'$$ (in mol/L). Then, at equilibrium:
$$
\mathrm{[Ba^{2+}]}= 0.0200 + s'\approx 0.0200~\mathrm{M} \quad (\text{since } s'\ll0.0200)
$$
and
$$
\mathrm{[CrO_{4}^{2-}]}= s'.
$$

The $$K_{sp}$$ expression becomes:
$$
K_{sp}= (0.0200)(s').
$$
Thus,
$$
s'=\frac{K_{sp}}{0.0200}=\frac{1.17\times10^{-10}}{0.0200}\approx 5.85\times10^{-9}~\mathrm{M}.
$$

Convert $$s'$$ to grams per liter:
$$
\text{Solubility (g/L)} = s' \times M_{\mathrm{BaCrO}_{4}}\approx (5.85\times10^{-9}~\mathrm{mol/L})(253.33~\mathrm{g/mol})\approx 1.48\times10^{-6}~\mathrm{g/L}.
$$

Rounded to 2 significant digits:
$$
1.5\times10^{-6}~\mathrm{g/L}.
$$

---

**Final Answers**

- Solubility in pure water: $$2.7\times10^{-3}~\mathrm{g/L}$$.
- Solubility in $$0.0200~\mathrm{M}$$ $$\mathrm{BaCl}_{2}$$ solution: $$1.5\times10^{-6}~\mathrm{g/L}$$.
- Solubility in pure water: $$2.7 \times 10^{-3}~\frac{g}{L}$$. - Solubility in $$0.0200~\mathrm{M}$$ $$\mathrm{BaCl}_{2}$$ solution: $$1.5 \times 10^{-6}~\frac{g}{L}$$.

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