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One type of kidney stone is a precipitate of calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.3 \times 10^{-9}\right) .\) A urine sample has a \(\mathrm{Ca}^{2+}\) concentration of \(2.5 \times 10^{-3} \mathrm{M}\) and an oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\right.\), \(\left.K_{\mathrm{a} 1}=5.9 \times 10^{-2}, \quad K_{\mathrm{a} 2}=6.4 \times 10^{-5}\right)\) concentration of \(1.1 \times 10^{-4} \mathrm{M}\) (a) A typical \(\mathrm{pH}\) for urine is \(5.5 .\) Will a precipitate of calcium oxalate form under these conditions? (b) A vegetarian diet results in a higher \(\mathrm{pH}\) for urine, typically greater than 7 . Would kidney stones be more or less likely to form in urine with a higher \(\mathrm{pH}\) ?

Step by step solution

01

Determine Concentration of Oxalate Ions

Oxalic acid \((\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4)\) is a diprotic acid and will dissociate into oxalate ions. The given dissociation constants are \(K_{\mathrm{a} 1}=5.9 \times 10^{-2}\) and \(K_{\mathrm{a} 2}=6.4 \times 10^{-5}\). Using these, calculate the oxalate ion concentration: \([\mathrm{C}_2\mathrm{O}_4^{2-}]\).

02

Calculate [HC鈧侽鈧勨伝] and [C鈧侽鈧劼测伝] at pH 5.5

For the given urine \(\mathrm{pH}=5.5\), calculate the concentration of \(\mathrm{HC}_2\mathrm{O}_4^-\) using \(K_{\mathrm{a}1}\) and approximate \(\mathrm{C}_2\mathrm{O}_4^{2-}\) using \(K_{\mathrm{a}2}\). Given \(\text{{initial } } \mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 = 1.1 \times 10^{-4}\), assume first ionization is complete: \ [\mathrm{HC}_2\mathrm{O}_4^-] \approx 1.1 \times 10^{-4} \, M\.

03

Calculate Ionic Product and Compare to Ksp

Calculate the ionic product, given \(\text{{[Ca}^{2+}]} = 2.5 \times 10^{-3}\, M\), using \ ([\text{{C}_2\mathrm{O}_4}^{2-}])\ approximated from the second dissociation: \ Q = [\mathrm{Ca}^{2+}][\mathrm{C}_2\mathrm{O}_4^{2-}]\. Compare \(Q\) to \(K_{\mathrm{sp}}\) of calcium oxalate (\(2.3 \times 10^{-9}\)).

04

Analyze Effect of Higher pH

At a \(\mathrm{pH} > 7\), the equilibrium favors \(\text{{C}_2\mathrm{O}_4}^{2-}\) more due to reduced \[\text{{H}^+}\], thus increasing \[\text{{C}_2\mathrm{O}_4}^{2-}\] and \[Q\]. Evaluate whether a higher \(\mathrm{pH}\) would increase or decrease the likelihood of precipitating calcium oxalate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kidney Stones Formation

Kidney stones are hard deposits that form in the kidneys. Importantly, one type of these stones is made up of calcium oxalate. They form when calcium ions and oxalate ions combine and surpass their solubility limit. This condition often becomes painful if a stone moves into the urinary tract. For calcium oxalate stones to form, the concentration of calcium ions \(\text{Ca}^{2+}\) and oxalate ions \(\text{C}_2\text{O}_4^{2-}\) exceeds the solubility product constant \(K_{sp}\). When the ionic product \(Q\) exceeds \(K_{sp}\), a precipitate forms. Understanding these conditions can help in predicting and preventing kidney stones.

pH Effect on Solubility

The pH level significantly affects the solubility of compounds in solution. For calcium oxalate, which precipitates from \(\text{Ca}^{2+}\) and \(\text{C}_2\text{O}_4^{2-}\) ions, the pH plays a critical role. At a lower pH, there are more hydrogen ions \(\text{H}^+\), which react with oxalate ions to form hydrogen oxalate \(\text{HC}_2\text{O}_4^-\). This reduces the availability of free \(\text{C}_2\text{O}_4^{2-}\) ions, thus decreasing the likelihood of calcium oxalate precipitating. In contrast, at a higher pH, there are fewer \(\text{H}^+\) ions, increasing \(\text{C}_2\text{O}_4^{2-}\) availability. This shift makes precipitation more likely, affecting kidney stone formation risk.

Ionic Product and Solubility Product

To determine if a precipitate forms, compare the ionic product \(Q\) with the solubility product constant \(K_{sp}\). The ionic product \(Q\) is found by multiplying the concentrations of the involved ions. A precipitate forms when \(Q\) exceeds \(K_{sp}\). For calcium oxalate, \(Q = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}]\). If \(Q\) is greater than \(2.3 \times 10^{-9}\), the compound is likely to precipitate out of the solution. Monitoring these values helps predict whether kidney stones will form and provides a basis for dietary and medicinal interventions.

Oxalic Acid Dissociation

Oxalic acid \(\text{H}_2\text{C}_2\text{O}_4\) is a diprotic acid, meaning it can lose two hydrogen ions \(\text{H}^+\). It dissociates in two steps, each with its own equilibrium constant \(K_a\). The first dissociation involves releasing one \(\text{H}^+\) to form hydrogen oxalate \(\text{HC}_2\text{O}_4^-\), with \(K_{a1} = 5.9 \times 10^{-2}\). The second dissociation releases another \(\text{H}^+\), forming oxalate \(\text{C}_2\text{O}_4^{2-}\), with \(K_{a2} = 6.4 \times 10^{-5}\). Knowing these constants helps estimate the concentration of \(\text{C}_2\text{O}_4^{2-}\) in a solution, which is crucial to understanding whether conditions are favorable for calcium oxalate precipitation.