A painful arthritic condition known as gout is caused by an excess of uric acid HUric in the blood. An aqueous solution contains \(4.00 \mathrm{~g}\) of uric acid. A \(0.730 \mathrm{M}\) solution of \(\mathrm{KOH}\) is used for titration. After \(12.00 \mathrm{~mL}\) of KOH is added, the resulting solution has pH 4.12. The equivalence point is reached after a total of \(32.62 \mathrm{~mL}\) of KOH is added. $$\operatorname{HUric}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Uric}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}$$ (a) What is the molar mass of uric acid? (b) What is its \(K_{\mathrm{a}}\) ?

Short Answer

Expert verified
Answer: The molar mass of uric acid is 168.07 g/mol and its Ka is 4.43 × 10^(-6).

Step by step solution

01

Find moles of KOH at the equivalence point

At the equivalence point, the number of moles of uric acid equals the number of moles of KOH. We know that 32.62 mL of 0.730 M KOH is needed to reach the equivalence point. So to find the moles of KOH: moles of KOH = Volume x Concentration = 32.62 mL × 0.730 mol/L = 0.0238 mol
02

Calculate the molar mass of uric acid

We know that 4.00 g of uric acid is in the solution, and from Step 1, we found that there are 0.0238 moles of uric acid. To find the molar mass of uric acid, divide the mass by the moles: Molar mass = Mass / Moles = 4.00 g / 0.0238 mol = 168.07 g/mol
03

Calculate the concentration of HUric at pH 4.12

We are given that, after adding 12.00 mL of KOH, the pH is 4.12. To find the concentration of HUric at this point, first, we need to convert pH to the hydronium ion concentration [H3O+]: [H3O+] = 10^(-pH) = 10^(-4.12) = 7.58 × 10^(-5) M
04

Calculate the moles of OH- added

We added 12.00 mL of 0.730 M KOH solution to reach this point in the titration. Calculate the moles of OH- ions added: moles of OH- = Volume × Concentration = 12.00 mL × 0.730 mol/L = 0.00876 mol
05

Find the concentration of Uric- and HUric ions

The reaction between HUric and OH- ions yields Uric- ions. Since we know the moles of OH- added (0.00876 mol), we can assume that the moles of Uric- ions formed are equal to the moles of OH- added. Also, since at this point in the titration, the moles of HUric have not changed, we can determine the moles of HUric remaining by subtracting the moles of OH- added from the initial moles of HUric (0.0238 mol): moles of HUric remaining = 0.0238 mol - 0.00876 mol = 0.0150 mol Now, we can find the concentration of Uric- and HUric ions by dividing their respective moles by the total volume of the solution (12.00 mL KOH added to 4.00 g uric acid solution, approximate 4.00 g as 4.00 mL): [Uric-] = 0.00876 mol / (4.00 mL + 12.00 mL) = 5.84 × 10^(-4) M [HUric] = 0.0150 mol / (4.00 mL + 12.00 mL) = 1.00 × 10^(-3) M
06

Calculate Ka for uric acid

We can use the formula for Ka, which is: Ka = ([Uric-][H3O+]) / [HUric] Plugging in the concentrations we calculated in Steps 3 and 5: Ka = (5.84 × 10^(-4) M × 7.58 × 10^(-5) M) / (1.00 × 10^(-3) M) = 4.43 × 10^(-6) So, the molar mass of uric acid is 168.07 g/mol and its Ka is 4.43 × 10^(-6).

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Key Concepts

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

Molar Mass Calculation
The concept of molar mass is critical when working with chemical substances because it links the mass of a substance to its moles, a foundational concept in chemistry. The molar mass of a compound is the weight in grams of one mole of that compound. In essence, it is the sum of the atomic masses of all the atoms in a molecule. To calculate it, you simply divide the total mass of the compound by the number of moles present.

Let's look at an example from a titration problem involving uric acid (HUric). If we know that 4.00 grams of uric acid corresponds to 0.0238 moles, the molar mass is calculated as follows: \[\text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}} = \frac{4.00\,\text{g}}{0.0238\,\text{mol}} = 168.07\,\text{g/mol}\]

This value is pivotal not only to understand the substance in question but also to make stoichiometric calculations in reactions, which is often the first step in many chemistry problems, such as determining the amount of reactants or products involved.
Acid Dissociation Constant (Ka)
The acid dissociation constant, usually denoted as Ka, measures the strength of an acid in a solution. It is the equilibrium constant for the dissociation of an acid into its conjugate base and a hydrogen ion (proton) in water. A large Ka value indicates a strong acid that dissociates completely in solution, while a small Ka value implies a weak acid that does not dissociate completely.

In our example with uric acid, we need to consider the pH of the solution, the point in the titration where we measured it, and the molar concentrations of the species in solution to calculate the Ka.

From the pH, we can determine the concentration of hydrogen ions (\[[H3O^+]\]) as follows:
\[[H3O^+] = 10^{-\text{pH}} = 10^{-4.12} = 7.58 \times 10^{-5} M\]

The concentrations of uric acid (HUric) and its conjugate base (Uric-) at specific titration points allow us to use the formula for Ka:
\[Ka = \frac{[\text{Uric}^-][H3O^+]}{[\text{HUric}]}\]
Substituting these values into the formula, we get the acid dissociation constant. The calculated Ka for uric acid is thus an expression of its tendency to lose a proton in aqueous solution.
pH and pOH Calculations
The pH and pOH are the logarithmic measures of hydrogen and hydroxide ion concentrations respectively. They are crucial for understanding the degree of acidity or basicity of a solution. The pH scale ranges from 0 to 14, with 7 being neutral. Values less than 7 indicate acidity, while values greater than 7 signify basicity.

pH is calculated using the concentration of hydrogen ions (\[[H3O^+]\]) in the solution which is inversely related to the pH value:\[\text{pH} = -\log[H3O^+]\]

Similarly, pOH is related to the hydroxide ions concentration \[[OH^-]\], and together, pH and pOH always add up to 14, a relationship derived from the ion-product constant of water \[K_w\]. To determine pH from a known concentration of a hydroxide ion, you first calculate the pOH and then subtract it from 14 to find the pH.

In a titration problem, the pH is especially significant because it changes markedly at the point where the amount of acid equals the amount of base added - known as the equivalence point. Understanding both pH and pOH is essential for analyzing acid-base titrations and for predicting the behavior of acidic and basic solutions in various chemical processes.

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Most popular questions from this chapter

WEB To make a buffer with \(\mathrm{pH}=3.0\) from \(\mathrm{HCHO}_{2}\) and \(\mathrm{CHO}_{2}^{-}\), (a) what must the [HCHO \(\left._{2}\right] /\left[\mathrm{CHO}_{2}^{-}\right]\) ratio be? (b) how many moles of \(\mathrm{HCHO}_{2}\) must be added to a liter of \(0.139 \mathrm{M}\) \(\mathrm{NaCHO}_{2}\) to give this \(\mathrm{pH}\) ? (c) how many grams of \(\mathrm{NaCHO}_{2}\) must be added to \(350.0 \mathrm{~mL}\) of \(0.159 \mathrm{MHCHO}_{2}\) to give this \(\mathrm{pH}\) ? (d) What volume of \(0.236 \mathrm{M} \mathrm{HCHO}_{2}\) must be added to \(1.00 \mathrm{~L}\) of a \(0.500 \mathrm{M}\) solution of \(\mathrm{NaCHO}_{2}\) to give this pH? (Assume that volumes are additive.)

WEB Write a balanced net ionic equation for the reaction of each of the following aqueous solutions with \(\mathrm{OH}^{-}\) ions. (a) ammonium nitrate (b) sodium dihydrogen phosphate \(\left(\mathrm{NaH}_{2} \mathrm{PO}_{4}\right)\) (c) \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\)

A \(0.4000 \mathrm{M}\) solution of nitric acid is used to titrate \(50.00 \mathrm{~mL}\) of \(0.237 \mathrm{M}\) barium hydroxide. (Assume that volumes are additive.) (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of nitric acid is required to reach the equivalence point? (d) What is the \(\mathrm{pH}\) of the solution before any \(\mathrm{HNO}_{3}\) is added? (e) What is the \(\mathrm{pH}\) of the solution halfway to the equivalence point? (f) What is the \(\mathrm{pH}\) of the solution at the equivalence point?

At \(25^{\circ} \mathrm{C}\) and \(1.00\) atm pressure, one liter of ammonia is bubbled into \(725 \mathrm{~mL}\) of water. Assume that all the ammonia dissolves and the volume of the solution is the volume of the water. A \(50.0-\mathrm{mL}\) portion of the prepared solution is titrated with \(0.2193 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) of the solution (a) before titration. (b) halfway to the equivalence point. (c) at the equivalence point.

A solution of an unknown weak acid at \(25^{\circ} \mathrm{C}\) has an osmotic pressure of \(0.878\) atm and a \(\mathrm{pH}\) of \(6.76 .\) What is \(K_{\mathrm{b}}\) for its conjugate base? (Assume that, in the equation for \(\pi\) [Chapter 10\(], i=1\).)

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