A painful arthritic condition known as gout is caused by an...

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 $\mathrm{KOH}$ are added, the resulting solution has $\mathrm{pH} 4.12 .$ The equivalence point is reached after a total of $32.62 \mathrm{~mL}$ of $\mathrm{KOH}$ are 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 $K_{\mathrm{a}}$ ?

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 $\mathrm{KOH}$
are added, the resulting solution has $\mathrm{pH} 4.12 .$ The equivalence point is reached after a total of $32.62 \mathrm{~mL}$ of $\mathrm{KOH}$ are 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 $K_{\mathrm{a}}$ ?

Key Concepts

Acid?Base Titration

Acid?base titration is an analytical technique used to determine the unknown concentration or amount of an acid or base by gradually adding a titrant of known concentration until the reaction reaches completion. This process relies on the neutralization reaction between the acid and the base, allowing the calculation of analyte properties based on volume and concentration data.

Stoichiometry of Titration

The stoichiometry in a titration refers to the quantitative relationship between the reacting acid and base. By knowing the mole ratio in the neutralization reaction, one can determine the number of moles of analyte and, consequently, other properties such as molar mass or concentration.

Equivalence Point

The equivalence point in a titration is the stage at which the amount of titrant added exactly equals the amount of substance present in the analyte based on the stoichiometric reaction. Reaching the equivalence point is critical for determining the total moles of the acid or base present.

Henderson?Hasselbalch Equation

The Henderson?Hasselbalch equation is used to relate the pH of a solution containing a weak acid and its conjugate base to the acid’s dissociation constant (Ka). This equation is essential for analyzing buffer solutions and for determining the acid strength during partial neutralization, where both the acid and its conjugate base are present.

Acid Dissociation Constant (Ka)

The acid dissociation constant, Ka, is a quantitative measure of the strength of a weak acid in solution. It is defined by the equilibrium concentration of the acid, its conjugate base, and the hydrogen ion. A higher Ka value indicates a stronger acid, while a lower value signifies a weaker acid.

Molar Mass Determination via Titration

Molar mass determination using titration data involves calculating the number of moles of a substance from the titration reaction and then relating it to the mass of the substance. By knowing the moles derived from the titrant volume and concentration, one can determine the molar mass of a compound by dividing its mass by the number of moles present.

Transcript

00:02 Okay, so we're given some information when we're at the equivalence point.

00:05 Okay, so when we're at the equivalence point here, we're given the amount of k -o -h.

00:13 So i'm going to use that to find its moles.

00:15 So i'm going to take the molarity of the k -o -h and multiply by its volume and liters, and see that it was 0 .2 -3 -8 moles of k -o -h that we added.

00:35 Okay, therefore that must have reacted with 0 .0 .238 moles of our uric acid.

00:47 Well, we were also told that there were four grams of our uric acid.

00:53 So molar mass is grams over moles.

00:58 So we'll take our four grams and divide by our moles.

01:05 0 .238 moles.

01:10 And we'll see that the molar mass of this uric acid is 168 grams per mole.

01:24 So before we hit the equivalence point, okay, we added 12 milliliters of k -o -h.

01:33 Since the equivalence point happened at 32 .62 milliliters, we know that we're before the equivalence point.

01:40 So we're going to end up with a buffer here.

01:43 Okay, i'll show you that, but we're going to end up with a buffer.

01:47 So first let's find the oh minus moles.

01:50 So we'll take the molarity of the k -o -h times it's volume and liters.

02:00 And we'll see that we added 0 .00876 moles of our oh minus.

02:11 And we had four grams of our uric acid.

02:19 Well, now that we have the molar mass, we can change grams to moles.

02:27 And we can see that we had 0 .238 moles of our uric acid.

02:40 So we're adding a strong base to a weak.

02:43 Acid.

02:45 So let's go ahead and write that equation and keep track of what's happening.

02:50 Okay, so here's our acid...

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