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Chapter 16: Problem 55

The molar solubility of \(\mathrm{MnCO}_{3}\) is \(4.2 \times 10^{-6} \mathrm{M}\). What is \(K_{\mathrm{sp}}\) for this compound?

Short Answer

Expert verified
The solubility product constant, \(K_{sp}\), for \(\mathrm{MnCO}_3\) is \(1.764 \times 10^{-11}\).

Step by step solution

01

Write the Dissolution Equation

\(\mathrm{MnCO}_{3}\) will dissociate in water according to the following reaction: \(\mathrm{MnCO}_3(s) \rightarrow \mathrm{Mn}^{2+}(aq) + \mathrm{CO}^{3-}_3(aq)\). In this case, one mole of \(\mathrm{MnCO}_3\) produces one mole each of \(\mathrm{Mn}^{2+}\) and \(\mathrm{CO}_3^{2-}\).
02

Expression for \(K_{sp}\)

The expression for \(K_{sp}\) is given by the product of the molar concentrations of the ions, each raised to the power equal to the coefficient in the balanced equation. So, \(K_{sp}=[\mathrm{Mn}^{2+}][\mathrm{CO}_3^{2-}]\). The \([ \]\) brackets denote molar concentrations at equilibrium.
03

Substitute Molar Solubility into \(K_{sp}\)

The molar solubility of \(\mathrm{MnCO}_3\) is \(4.2 \times 10^{-6} M\), this is the equilibrium concentration of both \(\mathrm{Mn}^{2+}\) and \(\mathrm{CO}_3^{2-}\) in solution. Substitute these concentrations into the \(K_{sp}\) expression: \(K_{sp}=(4.2 \times 10^{-6})(4.2 \times 10^{-6})\).
04

Calculate \(K_{sp}\)

Multiply the values to find \(K_{sp}\). The multiplication of \(4.2 \times 10^{-6}\) with itself results in \(1.764 \times 10^{-11}\).

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Most popular questions from this chapter

A student is asked to prepare a buffer solution at \(\mathrm{pH}=\) \(8.60,\) using one of the following weak acids: HA \(\left(K_{a}=2.7 \times 10^{-3}\right), \mathrm{HB}\left(K_{2}=4.4 \times 10^{-6}\right), \mathrm{HC}\left(K_{\mathrm{a}}=\right.\) \(2.6 \times 10^{-9}\) ). Which acid should she choose? Why?

A \(100-\mathrm{mL} 0.100 \mathrm{M} \mathrm{CuSO}_{4}\) solution is mixed with a \(100-\mathrm{mL} 0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution. Calculate the concentrations of the ions in the combined solution.

For which of the following reactions is the equilibrium constant called a solubility product? $$ \begin{array}{l} \text { (a) } \mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{OH}^{-}(a q) \rightleftharpoons \\ \text { (b) } 3 \mathrm{Ca}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons \mathrm{Ca}_{3}(\mathrm{OH})_{4}^{2-}(a q) \end{array} $$ (c) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightleftharpoons\) $$ \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) $$ (d) \(\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q)\)

How many milliliters of \(1.0 \mathrm{M} \mathrm{NaOH}\) must be added to a \(200 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaH}_{2} \mathrm{PO}_{4}\) to make a buffer solution with a pH of \(7.50 ?\)

Sketch titration curves for the following acid-base titrations: (a) \(\mathrm{HCl}\) versus \(\mathrm{NaOH},\) (b) \(\mathrm{HCl}\) versus \(\mathrm{CH}_{3} \mathrm{NH}_{2},\) (c) \(\mathrm{CH}_{3} \mathrm{COOH}\) versus \(\mathrm{NaOH}\). In each case, the base is added to the acid in an Erlenmeyer flask. Your graphs should show \(\mathrm{pH}\) on the \(y\) axis and volume of base added on the \(x\) axis.
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