A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured.
Mass of acid weighed out (grams) 0.732
Volume of NaOH required to reach endpoint: (ml) 17.8
pH of the mixture (half neutralized solution) 3.54
1) What is the molarity of the original acid solution?
2) What is the molecular weight of the acid?
Work:
1) (0.10 mmol / mL ) x (17.8 mL) = 1.78 mmol NaOH
1.78mmol*25 ml=44.5mmol Acid
44.5 M/50=.89-- This answer is wrong.
2).732 g /.0445 mol = 16.45 g/mol--Wrong
Many thanks for showing your work. It helps me spot the problem right off. BTW, I assume this is a monoprotic acid. Your error is in step 2. 1.78 mmols NaOH is correct. That means mmols HA is 1.78.
You took 25 mL for that titration and M = mmols/mL; therefore, 1.78 mmols/25 mL = ?M
For #2, mols = grams/molar mass or
molar mass = grams/mols. You had 0.732/2 grams titrated in the 25.00 mL and you had 0.0178 mols in that 25 mL; therefore, 0.365/0.0178 = about 20 molar mass.
One point I should bring up about the problem. Not only did the problem not state how many H ions were there, it also said the 25 mL was titrated to the neutral point. To me that means pH = 7.0 and if that's true I don't think all of the weak acid was titrated. But that's not part of the problem. I think it's just stated poorly.
To find the correct molarity of the original acid solution, you need to consider the volume used during the titration and the stoichiometry of the reaction.
1) Since you know that 17.8 mL of 0.10 M NaOH solution was required to reach the neutral endpoint, you can use this information to determine the number of moles of NaOH used.
0.10 M NaOH represents 0.10 moles of NaOH in 1 liter of solution, or 0.10 mmol in 1 mL of solution. Therefore, the number of moles of NaOH used in the titration is:
0.10 mmol/mL x 17.8 mL = 1.78 mmol NaOH
Since the stoichiometry of the reaction between the acid and NaOH is 1:1 (assuming the acid is monoprotic), this means that 1.78 mmol of the acid was also neutralized during the titration.
However, you need to take into account the dilution. The titrated portion was taken from the original 50.00 mL solution, so you only used half of it, which is 25.00 mL. This means that the 1.78 mmol of acid actually came from 25.00 mL of the original solution.
To find the molarity (M), you can use the formula:
Molarity = moles of solute / volume of solution (in liters)
Conversion: 25.00 mL = 25.00 mL x (1 L / 1000 mL) = 0.025 L
Molarity = 1.78 mmol / 0.025 L = 71.2 mmol/L = 0.0712 M
Hence, the molarity of the original acid solution is 0.0712 M.
Now, let's move on to the second question.
2) To find the molecular weight of the acid, you need to use the mass of the acid weighed out in grams and the number of moles of acid used during the titration.
The mass of the acid weighed out is given as 0.732 grams.
To find the number of moles (n) of the acid used, you can use the formula:
n = mass / molar mass
n = 0.732 g / molar mass
We already calculated that 44.5 mmol of acid was used during the titration. To convert this to moles:
44.5 mmol = 44.5 mmol x (1 mol / 1000 mmol) = 0.0445 mol
Now, we can set up the equation:
0.732 g / molar mass = 0.0445 mol
Rearranging the equation, we get:
molar mass = 0.732 g / 0.0445 mol
molar mass = 16.49 g/mol (rounded to two decimal places)
Therefore, the molecular weight of the acid is approximately 16.49 g/mol.
To determine the molarity of the original acid solution, you need to use the volume and concentration of the NaOH used in the titration. Here's how you can calculate it:
1) Calculate the number of moles of NaOH used:
Number of moles of NaOH = (0.10 mol/L) x (0.0178 L) = 0.00178 mol
2) Since you know that the acid is a weak acid and reacts with NaOH in a 1:1 ratio, the number of moles of the acid is also 0.00178 mol.
3) Now, calculate the molarity of the original acid solution:
Molarity of the original acid solution = (0.00178 mol) / (0.025 L) = 0.0712 M
So, the molarity of the original acid solution is 0.0712 M.
To calculate the molecular weight of the acid, you need to use the mass of the acid weighed out and the number of moles calculated above. Here's how:
1) Calculate the number of moles of acid:
Number of moles of acid = (0.732 g) / (molecular weight of acid)
2) Equate this to the number of moles calculated earlier:
(0.732 g) / (molecular weight of acid) = 0.00178 mol
3) Solve for the molecular weight of the acid:
Molecular weight of acid = (0.732 g) / (0.00178 mol)
So, the molecular weight of the acid is equal to (0.732 g) / (0.00178 mol).
Please note that the calculations provided above assume that the acid behaves as a monoprotic weak acid and reacts with NaOH in a 1:1 ratio.