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Chapter 17: Problem 102

Calculate the ratio of \(\left[\mathrm{Ca}^{2+}\right]\) to \(\left[\mathrm{Fe}^{2+}\right]\) in a lake in which the water is in equilibrium with deposits of both \(\mathrm{CaCO}_{3}\) and \(\mathrm{FeCO}_{3}\). Assume that the water is slightly basic and that the hydrolysis of the carbonate ion can therefore be ignored.

Short Answer

Expert verified
The ratio of the concentrations of Ca²⁺ and Fe²⁺ ions in the lake is approximately 101.5:1.

Step by step solution

01

Write the chemical equilibrium equations

The chemical equilibrium equations for the dissolution of CaCO₃ and FeCO₃ in water are as follows: \(CaCO_{3}(s) \rightleftharpoons Ca^{2+}(aq) + CO_{3}^{2-}(aq)\) \(FeCO_{3}(s) \rightleftharpoons Fe^{2+}(aq) + CO_{3}^{2-}(aq)\)
02

Write the solubility product expressions

The solubility product expressions (Ksp) for the above equilibrium equations would be: \(K_{sp(CaCO_{3})} = [Ca^{2+}][CO_{3}^{2-}]\) \(K_{sp(FeCO_{3})} = [Fe^{2+}][CO_{3}^{2-}]\)
03

Obtain the solubility product constants

The solubility product constants (Ksp) for CaCO₃ and FeCO₃ can be found in a reference book or table. For this problem, let's assume the following Ksp values: \(K_{sp(CaCO_{3})} = 3.36 \times 10^{-9}\) \(K_{sp(FeCO_{3})} = 3.31 \times 10^{-11}\)
04

Express both concentrations in terms of a common variable

Let's express the concentrations of both Ca²⁺ and Fe²⁺ in terms of the concentration of the carbonate ion (CO₃²⁻). Using the solubility product expressions from Step 2, we can write: \([Ca^{2+}] = \frac{K_{sp(CaCO_{3})}}{[CO_{3}^{2-}]}\) \([Fe^{2+}] = \frac{K_{sp(FeCO_{3})}}{[CO_{3}^{2-}]}\)
05

Calculate the ratio of the concentrations

Now we will find the ratio of the concentrations of Ca²⁺ and Fe²⁺ ions: \(\frac{[Ca^{2+}]}{[Fe^{2+}]} = \frac{\frac{K_{sp(CaCO_{3})}}{[CO_{3}^{2-}]}}{\frac{K_{sp(FeCO_{3})}}{[CO_{3}^{2-}]}} = \frac{K_{sp(CaCO_{3})}}{K_{sp(FeCO_{3})}}\) Using the Ksp values obtained in Step 3, we can calculate the ratio: \(\frac{[Ca^{2+}]}{[Fe^{2+}]} = \frac{3.36 \times 10^{-9}}{3.31 \times 10^{-11}} = 101.5\) Therefore, the ratio of the concentrations of Ca²⁺ and Fe²⁺ ions in the lake is approximately 101.5:1.

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Most popular questions from this chapter

A sample of \(7.5 \mathrm{~L}\) of \(\mathrm{NH}_{3}\) gas at \(22{ }^{\circ} \mathrm{C}\) and 735 torr is bubbled into a 0.50 - \(\mathrm{L}\) solution of \(0.40 \mathrm{M} \mathrm{HCl}\). Assuming that all the \(\mathrm{NH}_{3}\) dissolves and that the volume of the solution remains \(0.50 \mathrm{~L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

Suggest how the cations in each of the following solution mixtures can be separated: (a) \(\mathrm{Na}^{+}\) and \(\mathrm{Cd}^{2+},(\mathbf{b}) \mathrm{Cu}^{2+}\) and \(\mathrm{Mg}^{2+}\), (c) \(\mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+},(\mathbf{d}) \mathrm{Ag}^{+}\) and \(\mathrm{Hg}^{2+}\).

(a) Why is the concentration of undissolved solid not explicitly included in the expression for the solubility-product constant? (b) Write the expression for the solubility-product constant for each of the following strong electrolytes: AgI, \(\mathrm{SrSO}_{4}, \mathrm{Fe}(\mathrm{OH})_{2},\) and \(\mathrm{Hg}_{2} \mathrm{Br}_{2}\)

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{s p}=6.8 \times 10^{-27} .\) As discussed in the "Chemistry and Life" box on page 730 , fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F},\) whose \(K_{s p}=1.0 \times 10^{-60} \cdot(\mathrm{a})\) Write the expres- sion for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Predict whether the equivalence point of each of the following titrations is below, above, or at \(\mathrm{pH}\) 7: (a) \(\mathrm{NaHCO}_{3}\) titrated with \(\mathrm{NaOH},\) (b) \(\mathrm{NH}_{3}\) titrated with \(\mathrm{HCl}\) (c) KOH titrated with HBr.
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