Suppose that 0.483 g of an unknown monoprotic weak acid, HA, is dissolved in water. Titration of the solution with 0.250 M NaOH(aq) required 42.0 mL to reach the stoichiometric point. After the addition of 21.0 mL, the pH of the solution was found to be 3.75.
(a) What is the molar mass of the acid?
(b) What is the value of pKa for the acid?
mols acid = mols base
mols base = M x L = ?
Then mols acid = grams acid/molar mass. You know mols acid and grams, solve for molar mass.
Solve for mols base at 21.0 mL
Solve for mols acid remaining at 21.0 mL base
Substitute into H-H equation (see your post above) and solve for pKa.
Molar mass is 40.77g/mol
Pka=2.4×10M
To solve this problem, we need to use the concept of titration and the Henderson-Hasselbalch equation. Let's work through it step by step:
Step 1: Calculate the moles of acid used in the titration.
Given that the molarity of NaOH is 0.250 M, and the volume used is 42.0 mL, we can calculate the moles of NaOH used:
moles of NaOH = molarity × volume = 0.250 M × 0.0420 L = 0.0105 mol of NaOH
Since the acid and base react in a 1:1 ratio, the moles of acid used in the titration are also 0.0105 mol.
Step 2: Calculate the molar mass of the acid.
Given that 0.483 g of acid was dissolved, we can use the moles of acid calculated in Step 1 to find the molar mass:
molar mass = mass / moles = 0.483 g / 0.0105 mol ≈ 46.0 g/mol
Therefore, the molar mass of the acid is approximately 46.0 g/mol.
Step 3: Calculate the initial concentration of the acid.
Using the volume of NaOH solution added at the point where the pH was measured (21.0 mL) and assuming a 1:1 stoichiometry between acid and base, we can calculate the moles of acid neutralized:
moles of acid neutralized = moles of NaOH used = 0.0105 mol
Since the volume used is 21.0 mL, we convert it to liters:
volume = 21.0 mL = 0.0210 L
The initial concentration of the acid can be calculated as:
initial concentration = moles of acid neutralized / volume = 0.0105 mol / 0.0210 L ≈ 0.50 M
Therefore, the initial concentration of the acid is approximately 0.50 M.
Step 4: Calculate the pKa value.
Given that the pH of the solution is 3.75, we can use the Henderson-Hasselbalch equation to calculate the pKa value:
pH = pKa + log([A-] / [HA])
The [A-] / [HA] ratio can be approximated using the initial concentration of the acid:
[A-] / [HA] ≈ 10^(pH - pKa)
Solving for pKa, we have:
pKa = pH - log([A-] / [HA])
pKa = 3.75 - log(10^(3.75 - pKa))
Since the pH and pKa values are close, we can use an iterative approach to solve for pKa. Assuming an initial estimate of pKa = 4.0, let's calculate the ratio [A-] / [HA]:
[A-] / [HA] ≈ 10^(3.75 - 4.0) = 0.5623
Substituting this value into the equation, we get:
pKa = 3.75 - log(0.5623) ≈ 4.15
Therefore, the value of pKa for the acid is approximately 4.15.
To answer these questions, we need to use the concepts of titration and the Henderson-Hasselbalch equation. Here's how we can solve each part:
(a) To find the molar mass of the acid (HA), we can use the information about the mass of the acid and the volume of the NaOH solution needed to reach the stoichiometric point.
First, let's find the number of moles of NaOH used. Since the volume of NaOH solution used is 42.0 mL, we can convert it to liters by dividing by 1000: 42.0 mL / 1000 = 0.042 L.
The moles of NaOH used can be calculated using the equation:
moles NaOH = Molarity NaOH * Volume NaOH (in liters)
= 0.250 M * 0.042 L
= 0.0105 mol
Since the acid (HA) is monoprotic, the balanced equation for the neutralization reaction between HA and NaOH is:
HA(aq) + NaOH(aq) -> H2O(l) + NaA(aq)
From the stoichiometry, we know that the ratio of moles of HA to moles of NaOH is 1:1. Therefore, the number of moles of HA used is also 0.0105 mol.
Now, we can calculate the molar mass of HA by dividing the mass of HA used by the number of moles of HA:
molar mass of HA = mass of HA / moles of HA
= 0.483 g / 0.0105 mol
= 46.0 g/mol
Therefore, the molar mass of the acid (HA) is 46.0 g/mol.
(b) To find the value of pKa for the acid, we can use the pH value of the solution after the addition of 21.0 mL of NaOH and the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where [A-] is the concentration (in moles per liter) of the conjugate base (NaA) and [HA] is the concentration (in moles per liter) of the weak acid (HA).
Since the titration has reached the stoichiometric point after 42.0 mL (which means we have added half the volume needed for complete neutralization), the volume at the halfway point is 42.0 mL / 2 = 21.0 mL.
To calculate the concentration of NaA, we need to find the number of moles of NaA formed.
The moles of NaOH used can be calculated using the equation:
moles NaOH = Molarity NaOH * Volume NaOH (in liters)
= 0.250 M * 0.021 L
= 0.00525 mol
Since the ratio of NaOH to NaA is 1:1, we have also formed 0.00525 mol of NaA.
To find the concentration of NaA, we need to convert the volume to liters:
concentration of NaA = moles NaA / Volume NaA (in liters)
= 0.00525 mol / 0.021 L
= 0.25 M
Now we can substitute the values into the Henderson-Hasselbalch equation to solve for pKa:
pH = pKa + log([A-]/[HA])
3.75 = pKa + log(0.25/[HA])
Since [A-]/[HA] = [NaA]/[HA] = concentration of NaA / concentration of HA = 0.25 M / [HA], we can rearrange the equation:
pH - log(0.25/[HA]) = pKa
pKa = 3.75 - log(0.25/[HA])
Since we know the pH, we can substitute the values and solve for pKa.
Therefore, the value of pKa for the acid is 3.75 - log(0.25/[HA]).