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Chapter 6: Problem 90

The acid dissociation constant of uric acid is \(K_{a}=4.0 \times 10^{-6} M\). The \(\mathrm{pH}\) of a sample is \(6.0\). What is the ratio of urate ion to uric acid in the urine? (a) \(2.0\) (b) \(4.0\) (c) \(6.0\) (d) \(0.25\)

Short Answer

Expert verified
The ratio of urate ion to uric acid in the urine is 4.0.

Step by step solution

01

Calculate the hydronium ion concentration from pH

The pH of a solution is the negative logarithm of the hydronium ion concentration. To find the hydronium ion concentration, use the formula \[ [H_3O^+] = 10^{-\text{pH}} \]. Since the pH is given as 6.0, the hydronium ion concentration is \[ [H_3O^+] = 10^{-6.0} \text{ M} \].
02

Use the acid dissociation constant

Given the acid dissociation constant (\(K_a\)) of uric acid, set up the expression to relate the concentrations of urate ion, uric acid, and hydronium ion: \[ K_a = \frac{[\text{urate}^-]}{[\text{uric acid}] [H_3O^+]} \]. We will rearrange this to solve for the ratio \(\frac{[\text{urate}^-]}{[\text{uric acid}]}\).
03

Rearrange the equation to solve for the ratio

The equation from Step 2 can be rearranged to solve for the ratio \(\frac{[\text{urate}^-]}{[\text{uric acid}]}\): \[ \frac{[\text{urate}^-]}{[\text{uric acid}]} = \frac{K_a}{[H_3O^+]} \]. Substitute the known values to find the ratio. \[ \frac{[\text{urate}^-]}{[\text{uric acid}]} = \frac{4.0 \times 10^{-6}}{10^{-6}} \].
04

Calculate the ratio of urate ion to uric acid

Solve the equation by inserting the known values to find the ratio: \[ \frac{[\text{urate}^-]}{[\text{uric acid}]} = \frac{4.0 \times 10^{-6}}{10^{-6}} = 4.0 \].

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Key Concepts

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

pH Calculation
Understanding the pH of a solution is crucial for many chemical and biological processes. The pH is a scale used to measure the acidity or basicity of an aqueous solution. It is defined as the negative base 10 logarithm of the hydronium ion ([H3O+]) concentration. When the pH value is low, the concentration of hydronium ions is high, indicating an acidic solution. Conversely, a high pH value corresponds to a low concentration of hydronium ions, indicating a basic solution.

To calculate the pH, you can use the formula

3O+])

In the given exercise, the pH is 6.0, which means the hydronium ion concentration is [p>[H3O+] = 10-6.0 M

It is important to remember that the pH scale is logarithmic, which means each whole pH value below 7 is ten times more acidic than the next higher value. For example, a pH of 3 is ten times more acidic than a pH of 4 and a hundred times more acidic than a pH of 5. This concept is a key fundamental in understanding the behavior of acids and bases in solution.
Hydronium Ion Concentration
The hydronium ion ([H3O+]) concentration in a solution directly affects its pH level. Accurately determining this concentration is vital when dealing with acid-base chemistry, as it helps to understand the strength and nature of acids and bases in a given solution. Hydronium ions are formed when an acid dissociates in water and donates a proton (H+) to a water molecule.

The concentration of hydronium ions is calculated by taking the inverse logarithm of the pH value, using the expression:

[H3O+] = 10-pH

In our exercise, this calculation led to a hydronium ion concentration of 10-6.0 M, derived from the given pH of 6.0. It's critical to grasp that small changes in hydronium ion concentration can lead to significant shifts in pH, due to the logarithmic nature of the pH scale, as previously mentioned.
Urate to Uric Acid Ratio
The ration between urate ions and uric acid in a solution provides insight into the state of equilibrium in the context of acid dissociation. In our exercise, we are dealing with the acid dissociation constant (Ka) of uric acid. The Ka value is a measurement of the strength of an acid in solution and is defined for the dissociation of an acid (HA) into its conjugate base (A-) and hydronium ions (H3O+), as follows:

Ka = [A-]/[HA][H3O+]

Calculating the ration of urate (A-) to uric acid (HA) involves rearranging the expression to solve for [A-]/[HA]. By knowing the Ka value and the [H3O+] concentration, we can determine this ratio to understand how much uric acid has dissociated into urate ion within a sample.

When applying the given values from the exercise (Ka= 4.0 x 10-6 M and [H3O+]= 10-6 M), the ration of urate to uric acid is calculated to be 4.0. This indicates that, at a pH of 6.0, there are four times as many urate ions as there are uric acid molecules, revealing a higher degree of ionization of uric acid under these conditions. This kind of analysis is essential in biochemistry, specifically related to the metabolism of nucleic acids and the understanding of certain disorders like gout, where uric acid levels become clinically relevant.

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Most popular questions from this chapter

When equal volumes of the following solutions are mixed, the precipitation of AgCI \(\left(K_{s p}=1.8 \times 10^{-10}\right)\) will occur with : \(\begin{array}{ll}\text { (a) } 10^{-4} M\left(\mathrm{Ag}^{+}\right) \text {and } 10^{-4} M\left(\mathrm{Cl}^{-}\right) & \text {(b) } 10^{-5} M\left(\mathrm{Ag}^{+}\right) \text {and } 10^{-5} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\end{array}\) (c) \(10^{-5} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\) and \(10^{-6} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (d) \(10^{-10} M\left(\mathrm{Ag}^{+}\right)\) and \(10^{-10} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\)

The solubility of \(\mathrm{AgBrO}_{3}\) (formula weight \(=236\) ) is \(0.0072 \mathrm{~g}\) in \(1000 \mathrm{~mL}\). What is the \(K_{s p} ?\) (a) \(2.2 \times 10^{-8}\) (b) \(3.0 \times 10^{-10}\) (c) \(3.0 \times 10^{-5}\) (d) \(9.3 \times 10^{-10}\)

What is \(\left[\mathrm{NH}_{4}^{+}\right]\) in a solution that contain \(0.02 \mathrm{M} \mathrm{NH}_{3}\left(K_{b}=1.8 \times 10^{-5}\right)\) and \(0.01 \mathrm{M}\) ' \(\mathrm{KOH}\) ? (a) \(9 \times 10^{-6}\) (b) \(1.8 \times 10^{-5}\) (c) \(3.6 \times 10^{-5}\) (d) None of these

\(1.0 L\) solution is prepared by mixing \(61 \mathrm{gm}\) benzoic acid \(\left(p K_{a}=4.2\right)\) with \(72 \mathrm{gm}\) of sodium benzoate and then \(300 \mathrm{~mL} 1.0 \mathrm{M}\) HBr solution was added. The \(\mathrm{pH}\) of final solution is: (a) \(3.6\) (b) \(3.8\) (c) \(4.2\) (d) \(4.8\)

From separate solutions of four sodium salts NaW, NaX, NaY and \(\mathrm{NaZ}\) had \(\mathrm{pH} 7.0,9.0,10.0\) and \(11.0\) respectively, when each solution was \(0.1 M\), the weakest acid is : (a) HW (b) \(\mathrm{HX}\) (c) HY. (d) \(\mathbf{H Z}\)
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