a 3.54 grams solid sample of an unknown monoprotic acid was dissolved in distilled water to produce a 47.0 mL solution at 25 degrees. This solution was then titrated with 0.2 M NaOH. The equivalence point was reached when 35.72 mL of 0.2 M NaOH was delivered.
A. find the number of moles of acid in the solid sample.
B. Find the molar mass of the unknown acid
C. Find Ka for the unknown acid if the pH at the equivalence point is 8.79
D. What was the pH after 63.2 mL of 0.2 M NaOH was delivered?
Can you tell me what the main points troubling you? Please explain in detail.
finding all the answers
4,67
A. To find the number of moles of acid in the solid sample, we first need to calculate the number of moles of NaOH used at the equivalence point.
The number of moles of NaOH used can be calculated using the equation:
moles of NaOH = concentration of NaOH * volume of NaOH
Given that the concentration of NaOH is 0.2 M and the volume of NaOH used is 35.72 mL (or 0.03572 L), we have:
moles of NaOH = 0.2 M * 0.03572 L = 0.007144 mol
Since NaOH reacts in a 1:1 ratio with the monoprotic acid, the number of moles of acid in the solid sample is also 0.007144 mol.
B. The molar mass of the unknown acid can be calculated using the number of moles of acid and the mass of the solid sample.
The molar mass of the unknown acid is given by the equation:
molar mass = mass of solid sample / moles of acid
Given that the mass of the solid sample is 3.54 g and the moles of acid is 0.007144 mol, we have:
molar mass = 3.54 g / 0.007144 mol = 494.37 g/mol
Therefore, the molar mass of the unknown acid is 494.37 g/mol.
C. To find Ka for the unknown acid, we need to determine the concentration of the acid at the equivalence point. The concentration of the acid can be calculated using the volume of NaOH used at the equivalence point and the known concentration of NaOH.
First, we need to convert the volume of NaOH used to liters:
volume of NaOH = 35.72 mL = 0.03572 L
Since NaOH and the unknown acid react in a 1:1 ratio, the number of moles of the unknown acid is also 0.007144 mol.
The concentration of the unknown acid, [HA], can be calculated using the equation:
[HA] = moles of acid / volume of solution
Given that the volume of solution is 47.0 mL and the moles of acid is 0.007144 mol, we have:
[HA] = 0.007144 mol / 0.0470 L = 0.152 mol/L
Now, to find Ka for the unknown acid, we can use the equation:
Ka = [H+][A-] / [HA]
Since the pH at the equivalence point is given as 8.79, we know that [H+] = 10^(-pH). Therefore:
[H+] = 10^(-8.79) = 1.32 * 10^(-9) M
Substituting the values into the formula for Ka:
Ka = (1.32 * 10^(-9))(0.152) / 0.152
Ka = 1.32 * 10^(-9)
Therefore, the Ka value for the unknown acid is 1.32 * 10^(-9).
D. To calculate the pH after 63.2 mL of 0.2 M NaOH is delivered, we need to determine the moles of excess NaOH and the resulting concentration of OH- ions.
First, calculate the moles of NaOH in 63.2 mL (0.0632 L):
moles of NaOH = 0.2 M * 0.0632 L = 0.01264 mol
Since NaOH and H+ ions react in a 1:1 ratio, the moles of H+ ions neutralized by the excess NaOH is also 0.01264 mol.
The remaining moles of H+ ions can be calculated by subtracting the moles of H+ ions initially present (0.007144 mol) and the moles neutralized by the excess NaOH (0.01264 mol) from the moles of acid initially present (0.007144 mol):
moles of remaining H+ ions = 0.007144 mol - 0.01264 mol + 0.007144 mol = 0.001648 mol
Since the concentration of H+ ions is equal to the concentration of OH- ions at the equivalence point, the concentration of OH- ions is:
[OH-] = moles of remaining H+ ions / volume of solution
[OH-] = 0.001648 mol / 0.0470 L = 0.0350 mol/L
Now, we can calculate the pOH using the equation:
pOH = -log10[OH-] = -log10(0.0350) = 1.46
Finally, the pH can be calculated using the equation:
pH = 14 - pOH = 14 - 1.46 = 12.54
Therefore, the pH after 63.2 mL of 0.2 M NaOH is delivered is 12.54.