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Chapter 19: Problem 110

Gout is caused by an error in metabolism that leads to a buildup of uric acid in body fluids, which is deposited as slightly soluble sodium urate \(\left(\mathrm{C}_{5} \mathrm{H}_{3} \mathrm{~N}_{4} \mathrm{O}_{3} \mathrm{Na}\right)\) in the joints. If the extracellular \(\left[\mathrm{Na}^{+}\right]\) is \(0.15 \mathrm{M}\) and the solubility of sodium urate is \(0.085 \mathrm{~g} / 100 . \mathrm{mL}\) what is the minimum urate ion concentration (abbreviated [Ur-]) that will cause a deposit of sodium urate?

Short Answer

Expert verified
Minimum urate ion concentration is \(4.47 \times 10^{-3} \text{ M}\).

Step by step solution

01

Write the Dissociation Reaction

Sodium urate \(\text{C}_{5}\text{H}_{3}\text{N}_{4}\text{O}_{3}\text{Na}\) dissociates in water. The dissociation reaction is: \[\text{C}_{5}\text{H}_{3}\text{N}_{4}\text{O}_{3}\text{Na} \rightarrow \text{C}_{5}\text{H}_{3}\text{N}_{4}\text{O}_{3}^{-} + \text{Na}^{+}\]
02

Solubility Product Constant (Ksp)

The solubility product constant for sodium urate can be expressed as: \[\text{K}_{\text{sp}} = [\text{Na}^{+}][\text{C}_{5}\text{H}_{3}\text{N}_{4}\text{O}_{3}^{-}]\]
03

Calculate Molar Mass

Calculate the molar mass of \(\text{C}_{5}\text{H}_{3}\text{N}_{4}\text{O}_{3}\text{Na}\): \[\text{C}: 5 \times 12.01 = 60.05\ \text{H}: 3 \times 1.01 = 3.03\ \text{N}: 4 \times 14.01 = 56.04\ \text{O}: 3 \times 16.00 = 48.00\ \text{Na}: 1 \times 22.99 = 22.99\ \text{Total} = 190.11 \text{ g/mol}\]
04

Convert Solubility to Molarity

Given solubility is \[0.085 \text{ g} / 100 \text{ mL}\]. Convert this to molarity: \[\frac{0.085 \text{ g}}{100 \text{ mL}} \times \frac{1 \text{ mol}}{190.11 \text{ g}} \times \frac{1000 \text{ mL}}{1 \text{ L}} = 0.00447 \text{ M}\]
05

Substitute Values into Ksp Expression

Using the solubility product \(\text{K}_{\text{sp}}\) expression: \[\text{K}_{\text{sp}} = (0.15 \text{ M}) (0.00447 \text{ M}) = 6.705 \times 10^{-4}\]
06

Solve for Minimum Urate Ion Concentration

The minimum concentration of urate ion \[\text{[Ur}^-\text{]}\] is: \[\text{[Ur}^-\text{]} \geq \frac{6.705 \times 10^{-4}}{0.15} = 4.47 \times 10^{-3} \text{ M}\]

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Key Concepts

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

Dissociation Reaction
When a compound splits into its ions in water, we call it a dissociation reaction. For sodium urate, the dissociation reaction looks like this: \[ \text{C}_{5}\text{H}_{3}\text{N}_{4}\text{O}_{3}\text{Na} \rightarrow \text{C}_{5}\text{H}_{3}\text{N}_{4}\text{O}_{3}^{-} + \text{Na}^{+} \] This equation shows that one molecule of sodium urate dissociates into one sodium ion and one urate ion. This understanding is crucial for solving problems related to solubility and ion concentrations. It is the basis for calculating the solubility product constant (Ksp) later.
Solubility Product (Ksp)
The solubility product constant, commonly known as \( K_{sp} \), is a special constant used to describe the equilibrium between a solid and its respective ions in a solution. For sodium urate, the equation is: \[ K_{sp} = [\text{Na}^{+}] [\text{C}_{5}\text{H}_{3}\text{N}_{4}\text{O}_{3}^{-}] \] Here, the Ksp helps us understand the max concentrations of ions in the solution before it starts precipitating out as a solid. Knowing the molarity of these ions allows us to determine when we reach this equilibrium.
Molarity Conversion
To work with concentrations, we often need to convert between different units. In this case, we start with the solubility of sodium urate given in grams per 100 mL. Let's convert this to molarity: \[ \frac{0.085\ \text{g}}{100\ \text{mL}} = 0.00085\ \text{g/mL} \] Now, we find the molar mass of sodium urate, which equals 190.11 g/mol. The molarity is then: \[ \frac{0.00085\ \text{g/mL}}{190.11\ \text{g/mol}} \times 1000 = 0.00447\ \text{M} \] This conversion is significant because it allows us to insert the value into the Ksp equation.
Minimum Ion Concentration
To find the minimum concentration of urate ions that would cause sodium urate to start precipitating, we need to utilize the given Ksp value and the sodium ion concentration. By substituting values into the Ksp expression: \[ K_{sp} = [\text{Na}^{+}] [\text{C}_{5}\text{H}_{3}\text{N}_{4}\text{O}_{3}^{-}] \] \[ 6.705 \times 10^{-4} = )(0.15\ \text{M})([Ur^-]) \] Solving for \[ [Ur^-] \], we get: \[ [Ur^-] eq \frac{6.705 \times 10^{-4}}{0.15} = 4.47 \times 10^{-3}\ \text{M} \] This means urate ions will start precipitating out if their concentration is at least 4.47 mM.
Gout and Uric Acid
Gout results from an excess of uric acid in bodily fluids, leading to sodium urate deposition in joints. Understanding the solubility product and dissociation reactions helps medically manage and predict these deposits. Sodium urate, a slightly soluble compound, forms crystals that cause joint pain and inflammation when uric acid concentration exceeds the solubility limit. Comprehending molarity and the minimum ion concentration aids in identifying when these deposits can occur, thereby helping in the treatment and prevention of gout through dietary and pharmaceutical measures.

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Most popular questions from this chapter

What is the purpose of an acid-base buffer?

What is the component concentration ratio, \(\left[\mathrm{NO}_{2}^{-}\right] /\left[\mathrm{HNO}_{2}\right],\) of a buffer that has a \(\mathrm{pH}\) of \(2.95\left(K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HNO}_{2}=7.1 \times 10^{-4}\right) ?\)

Find the \(\mathrm{pH}\) of a buffer that consists of \(0.45 \mathrm{M} \mathrm{HCOOH}\) and \(0.63 M \mathrm{HCOONa}\left(\mathrm{p} K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HCOOH}=3.74\right)\)

A \(50.0-\mathrm{mL}\) volume of \(0.50 M \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) is mixed with \(125 \mathrm{~mL}\) of \(0.25 M \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) (a) If aqueous \(\mathrm{NaOH}\) is added, which ion precipitates first? (See Appendix C.) (b) Describe how the metal ions can be separated using \(\mathrm{NaOH}\). (c) Calculate the \(\left[\mathrm{OH}^{-}\right]\) that will accomplish the separation.

An ecobotanist separates the components of a tropical bark extract by chromatography. She discovers a large proportion of quinidine, a dextrorotatory isomer of quinine used for control of arrhythmic heartbeat. Quinidine has two basic nitrogens \(\left(K_{\mathrm{b} 1}=4.0 \times 10^{-6}\right.\) and \(\left.K_{\mathrm{b} 2}=1.0 \times 10^{-\mathrm{i} 0}\right) .\) To measure the concentration of quinidine, she carries out a titration. Because of the low solubility of quinidine, she first protonates both nitrogens with excess \(\mathrm{HCl}\) and titrates the acidified solution with standardized base. A 33.85-mg sample of quinidine \((\mathscr{M}=324.41 \mathrm{~g} / \mathrm{mol})\) is acidified with \(6.55 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{HCl}\) (a) How many milliliters of \(0.0133 \mathrm{M} \mathrm{NaOH}\) are needed to titrate the excess HCl? (b) How many additional milliliters of titrant are needed to reach the first equivalence point of quinidine dihydrochloride? (c) What is the \(\mathrm{pH}\) at the first equivalence point?
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