/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 97 Calculate the ratio of \(\left[\... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 97

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 [\(\mathrm{Ca}^{2+}\)] to [\(\mathrm{Fe}^{2+}\)] in the lake is approximately 101.5, as calculated by dividing the solubility product constants for CaCO鈧� and FeCO鈧�.

Step by step solution

01

Write the equilibrium reactions for both CaCO鈧� and FeCO鈧�

CaCO鈧�(s) 鈬� Ca虏鈦�(aq) + CO鈧兟测伝(aq) FeCO鈧�(s) 鈬� Fe虏鈦�(aq) + CO鈧兟测伝(aq)
02

Write the equilibrium expressions for both CaCO鈧� and FeCO鈧�

\(K_{sp}^{CaCO_3} = [\mathrm{Ca}^{2+}][\mathrm{CO_3}^{2-}]\) \(K_{sp}^{FeCO_3} = [\mathrm{Fe}^{2+}][\mathrm{CO_3}^{2-}]\)
03

Use the given information to find the ratio of [\(\mathrm{Ca}^{2+}\)] to [\(\mathrm{Fe}^{2+}\)]

Since the water is in equilibrium with both CaCO鈧� and FeCO鈧� deposits, the concentration of CO鈧兟测伝 ions will be the same in both cases. Therefore, we can divide the equilibrium expressions for both salts to find the ratio of concentrations of Ca虏鈦� and Fe虏鈦� ions: \(\frac{[\mathrm{Ca}^{2+}]}{[\mathrm{Fe}^{2+}]} = \frac{K_{sp}^{CaCO_3}}{K_{sp}^{FeCO_3}}\)
04

Find the values of solubility product constants for CaCO鈧� and FeCO鈧�

\(K_{sp}^{CaCO_3} = 3.36 \times 10^{-9}\) (from a table or given information) \(K_{sp}^{FeCO_3} = 3.31 \times 10^{-11}\) (from a table or given information)
05

Calculate the ratio of [\(\mathrm{Ca}^{2+}\)] to [\(\mathrm{Fe}^{2+}\)]

\(\frac{[\mathrm{Ca}^{2+}]}{[\mathrm{Fe}^{2+}]} = \frac{3.36 \times 10^{-9}}{3.31 \times 10^{-11}} = 101.5\) The ratio of [\(\mathrm{Ca}^{2+}\)] to [\(\mathrm{Fe}^{2+}\)] in the lake is approximately 101.5.

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Most popular questions from this chapter

(a) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood of pH \(7.4\) ? (b) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in an exhausted marathon runner whose blood pH is 7.1?

(a) Calculate the percentionization of \(0.085 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .\) (b) Calculate the percent ionization of \(0.095 \mathrm{M}\) lactic acid in a solution containing \(0.0075 \mathrm{M}\) sodium lactate.

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Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00\) L of solution. Buffer A is prepared using \(1.00 \mathrm{~mol}\) each of formic acid and sodium formate. Buffer \(B\) is prepared by using \(0.010\) mol of each. (a) Calculate the pH of each buffer, and explain why they are equal. (b) Which buffer will have the greater buffer capacity? Explain. (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 M \mathrm{HCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (e) Discuss your answers for parts (c) and (d) in light of your response to part (b).
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