Question

Kidney stones are caused by the precipitation of either calcium oxalate, Ca(C2O4), or calcium phosphate, Ca3(PO4)2, in the kidneys. If the normal concentration of Ca2+ in the kidneys is 2.5 mM, at what concentration of oxalate ion will kidney stones begin to form?

Answer 1

Answer:

1.1 × 10⁻⁶ M

Explanation:

Step 1: Given and required data

• Concentration of Ca²⁺ ([Ca²⁺]): 2.5 mM = 2.5 × 10⁻³ M

• Calcium oxalate solubility product constant (Ksp): 2.7 × 10⁻⁹

Step 2: Write the balanced equation for the solution reaction of calcium oxalate

CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

Step 3: Calculate the concentration of the oxalate ion to begin the precipitation of calcium oxalate

We will use the Ksp of calcium oxalate.

Ksp = 2.7 × 10⁻⁹ = [Ca²⁺].[C₂O₄²⁻]

[C₂O₄²⁻] = Ksp / [Ca²⁺]

[C₂O₄²⁻] = 2.7 × 10⁻⁹ / 2.5 × 10⁻³

[C₂O₄²⁻] = 1.1 × 10⁻⁶ M

Answer 2

Kidney stones are the calcareous stones formed in the kidney of the excretory system. The initial concentration of oxalate ion was $1.1 \times 10^{-6} \;\rm M.$

What is concentration?

Concentration is the amount of the substance present in the sample and can be calculated by the solubility constant of the products.

Given,

Concentration of calcium = 2.5mM

The solubility product constant = $2.7 \times 10^{-9}$

The balanced chemical reaction for calcium oxalate can be written as:

$\rm CaC_{2}O_{4}(s) \leftrightharpoons Ca^{2+}(aq) + C_{2}O_{4}^{2-}(aq)$

The concentration of oxalate ions by Ksp is calculated as:

$\begin{aligned} [C_{2}O_{4}^{2-}] &= \rm \dfrac {Ksp}{[Ca^{2+}]}\\\\&= \dfrac{2.7 \times 10^{-9}}{2.5 \times 10^{-3}}\\\\&= 1.1 \times 10^{-6}\;\rm M\end{aligned}$

Therefore, $1.1 \times 10^{-6} \;\rm M$ is the concentration of oxalate ion.

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