You weigh a sample of a monoprotic unknown acid and dissolve it in 50.00 mL of distilled water. Exactly half of this solution is titrated with Sodium Hydroxide to the phenolphthalein end point.
The pH of the other half of the original solution is measured with a pH meter. The "neutralized" solution is added to the "original" solution and the pH of this combined "final" solution is also measured.
The following are the measured values:
Mass of unknown acid 1.6214 g
Volume of NaOH used in titration 15.88 mL
Concentration of the NaOH used 0.1849 M
pH of the original acid solution 2.64
pH of the final acid solution 4.45
CALCULATE the following
(a) Molecular Weight of Acid used in titration ___________________
(b) Molarity of UNKNOWN Acid solution from titration ___________________
(c) Ka of UNKNOWN Acid ___________________
To calculate (a), the molecular weight of the acid used in titration, we can use the formula:
Molecular weight (MW) = Mass of acid / Moles of acid
To find the moles of acid, we need to use the data from the titration.
First, let's find the moles of sodium hydroxide (NaOH) used in the titration:
Moles of NaOH = Volume of NaOH used (in L) * Concentration of NaOH (in M)
= 15.88 mL * (1 L / 1000 mL) * 0.1849 M
= 0.002941 mol
Since the reaction between the acid and NaOH is 1:1, the moles of acid used in the titration are also 0.002941 mol.
Now, let's calculate the molecular weight of the acid:
MW = Mass of acid / Moles of acid
= 1.6214 g / 0.002941 mol
≈ 551.61 g/mol
Therefore, the molecular weight of the acid used in titration is approximately 551.61 g/mol.
To calculate (b), the molarity of the unknown acid solution from titration, we need to use the equation:
Molarity (M) = Moles of acid / Volume of acid solution (in L)
We already calculated the moles of acid used in the titration (0.002941 mol). Now we need to find the volume of the acid solution used in the titration.
Since the titrated solution is half of the original solution, the volume of the titrated acid solution is 50.00 mL / 2 = 25.00 mL = 0.025 L.
Now, let's calculate the molarity of the unknown acid solution:
Molarity = Moles of acid / Volume of acid solution
= 0.002941 mol / 0.025 L
= 0.1176 M
Therefore, the molarity of the unknown acid solution from titration is 0.1176 M.
To calculate (c), the Ka of the unknown acid, we can use the Henderson-Hasselbalch equation:
pH = pKa + log [A-] / [HA]
In this case, we know the pH of the original acid solution (2.64) and the pH of the final acid solution (4.45). We also know that at the phenolphthalein end point, all the acid is neutralized and converted to its conjugate base (A-). Therefore, in the final solution, the concentration of the acid (HA) and its conjugate base (A-) are equal.
Let's rearrange the Henderson-Hasselbalch equation to solve for pKa:
pKa = pH - log [A-] / [HA]
First, let's find the ratio [A-] / [HA]:
[A-] / [HA] = 10^(pH - pKa)
Using the pH and pKa values, we can calculate the ratio [A-] / [HA] for the final solution:
[A-] / [HA] = 10^(4.45 - 2.64)
= 10^(1.81)
≈ 63.28
Since [A-] equals [HA] in the final solution, we can assume that the concentration of the acid (HA) is 63.28 units.
Now, let's find the moles of the acid (HA) in the final solution:
Moles of acid (HA) = Concentration of acid (HA) * Volume of acid solution (in L)
= 63.28 * 0.025 L
= 1.582 mol
Since the moles of acid (HA) and its conjugate base (A-) are equal, the moles of A- are also 1.582 mol.
Finally, let's calculate the Ka of the unknown acid using the equation:
Ka = [H3O+] * [A-] / [HA]
At acidic pH, [H3O+] is approximately equal to 10^(-pH), so:
Ka = 10^(-pH) * [A-] / [HA]
= 10^(-4.45) * 1.582 / 1.582
≈ 3.16 * 10^(-5)
Therefore, the Ka of the unknown acid is approximately 3.16 * 10^(-5).