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Chapter 17: Problem 55

A 1.50-L solution saturated at \(25^{\circ} \mathrm{C}\) with cobalt carbonate \(\left(\mathrm{CoCO}_{3}\right)\) contains \(2.71 \mathrm{mg}\) of \(\mathrm{CoCO}_{3} .\) Calculate the solubility-product constant for this salt at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The solubility product constant for cobalt carbonate (CoCO3) at 25°C, calculated from a 1.50 L solution saturated with 2.71 mg of CoCO3, is \( K_{sp} = 2.314 \times 10^{-10} \).

Step by step solution

01

Convert the mass of CoCO3 to moles

We are given that the solution contains 2.71 mg of CoCO3. We need to convert this mass into moles. To do this, we'll use the molar mass of CoCO3, which can be calculated by adding the molar masses of its individual components: Co (58.93 g/mol), C (12.01 g/mol), and O3 (3 × 16.00 g/mol). The molar mass of CoCO3 is: CoCO3 = 58.93 + 12.01 + (3 × 16.00) = 118.94 g/mol Now we can convert the mass of CoCO3 into moles: moles of CoCO3 = mass of CoCO3 / molar mass of CoCO3 moles of CoCO3 = 2.71 mg × (1 g / 1000 mg) × (1 mol / 118.94 g) moles of CoCO3 = 2.282 × 10^{-5} mol
02

Calculate the molar concentration of CoCO3

We are given that the volume of the solution is 1.50 L. To find the molar concentration of CoCO3, we will divide the moles of CoCO3 by the volume of the solution in liters: [CoCO3] = moles of CoCO3 / volume of solution [CoCO3] = 2.282 × 10^{-5} mol / 1.50 L [CoCO3] = 1.522 × 10^{-5} M
03

Calculate the solubility product constant, Ksp

Now we can use the molar concentration of CoCO3 to calculate its solubility product constant, Ksp. The solubility product constant is given by: Ksp = [Co²⁺][CO3²⁻] Since one mole of CoCO3 dissociates into one mole of Co²⁺ and one mole of CO3²⁻, their molar concentrations will be equal. [Co²⁺] = [CO3²⁻] = 1.522 × 10^{-5} M Now we can calculate Ksp: Ksp = (1.522 × 10^{-5})(1.522 × 10^{-5}) Ksp = 2.314 × 10^{-10} The solubility product constant for cobalt carbonate (CoCO3) at 25°C is 2.314 × 10^{-10}.

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Most popular questions from this chapter

Compare the titration of a strong, monoprotic acid with a strong base to the titration of a weak, monoprotic acid with a strong base. Assume the strong and weak acid solutions initially have the same concentrations. Indicate whether the following statements are true or false. (a) More base is required to reach the equivalence point for the strong acid than the weak acid. \((\mathbf{b})\) The \(\mathrm{pH}\) at the beginning of the titration is lower for the weak acid than the strong acid. \((\mathbf{c})\) The \(\mathrm{pH}\) at the equivalence point is 7 no matter which acid is titrated.

A buffer contains 0.20 mol of acetic acid and 0.25 mol of sodium acetate in \(2.50 \mathrm{~L}\). (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.05 mol of \(\mathrm{NaOH}\) ? (c) What is the pH of the buffer after the addition of \(0.05 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is $2.5 \times 10^{-14} .(\mathbf{a})$ What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?(\mathbf{b})\) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to $1.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?$

(a) A 0.1044-g sample of an unknown monoprotic acid requires $22.10 \mathrm{~mL}\( of \)0.0500 \mathrm{M} \mathrm{NaOH}$ to reach the end point. What is the molar mass of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid.

Consider the titration of \(30.0 \mathrm{~mL}\) of $0.050 \mathrm{M} \mathrm{NH}_{3}\( with \)0.025 M$ HCl. Calculate the pH after the following volumes of titrant have been added: $(\mathbf{a}) 0 \mathrm{~mL},(\mathbf{b}) 20.0 \mathrm{~mL},\( (c) 59.0 \)\mathrm{mL},(\mathbf{d}) 60.0 \mathrm{~mL},(\mathbf{e}) 61.0 \mathrm{~mL},(\mathbf{f}) 65.0 \mathrm{~mL} .$
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