In the titration of total acid with sodium hydroxide, NaOH, it was determined that 36.03mL of 0.0980M NaOH was needed to neutralize all the acid in a 10.0mL aliquot of a powdered drink mix solution. In a second titration, it was determined that were 1.45 x 10^(-5) moles of ascorbic acid in a 10.0mL aliquot of the same powdered drink mix solution. Calculate the moles of citric acid in the 10.0mL aliquot of powdered drink mix solution.
So confused.. Thank you.
I believe ascorbic acid is a diprotic acid which I will label as H2A.
Total mols NaOH used = M x L = approx 0.00353 mols NaOH.
The solution contained 1.45E-5 mols ascorbic and that will use 2*1.45E-5 = 2.90E-5 mols NaOH.
That leaves 0.003531-2.90E-5 = 0.00350 for mols NaOH to titrate the citric acid in the 10.0 mL sample. Since citric acid is a triprotic acid, mols citric acid will be 1/3 mols NaOH = 0.00350/3 = ?
Check my work.
If you had to make a pitcher of punch with a powdered drink mix, explain how this lab would allow making this drink in the least amount of time
To calculate the moles of citric acid in the 10.0 mL aliquot of powdered drink mix solution, we can use the information from the titration of total acid with sodium hydroxide.
First, let's calculate the moles of NaOH used in the titration:
Moles of NaOH = (volume of NaOH used in liters) x (molarity of NaOH)
= 0.03603 L x 0.0980 mol/L
= 0.00353454 mol
Next, we need to determine the moles of ascorbic acid in the 10.0 mL aliquot of the powdered drink mix solution:
Moles of ascorbic acid = 1.45 x 10^(-5) mol
Since we know that citric acid is the only other acid present in the powdered drink mix solution, and assuming that citric acid and ascorbic acid react in a 1:1 ratio during the titration, the moles of citric acid in the 10.0 mL aliquot of the powdered drink mix solution is also 1.45 x 10^(-5) mol.
Therefore, the moles of citric acid in the 10.0 mL aliquot of the powdered drink mix solution is 1.45 x 10^(-5) mol.
To calculate the moles of citric acid in the 10.0 mL aliquot of powdered drink mix solution, we need to use the information given in the problem and apply stoichiometry.
First, let's analyze the titration of total acid with sodium hydroxide. We know that 36.03 mL of 0.0980 M NaOH was needed to neutralize all the acid in a 10.0 mL aliquot of the powdered drink mix solution.
From the reaction equation between the acid and NaOH, we can see that the ratio between the moles of acid and moles of NaOH is 1:1. This means that the moles of acid in the 10.0 mL aliquot is the same as the moles of NaOH used in the titration.
Using the formula for moles, we can calculate the moles of NaOH used:
Moles NaOH = Volume NaOH (in L) x Molarity NaOH
Converting the volume of NaOH used in the titration to liters:
36.03 mL = 0.03603 L
Calculating the moles of NaOH used:
Moles NaOH = 0.03603 L x 0.0980 M = 0.00353094 moles
Since the ratio between the moles of acid and moles of NaOH is 1:1, the moles of acid in the 10.0 mL aliquot is also 0.00353094 moles.
Next, we need to calculate the moles of ascorbic acid in the 10.0 mL aliquot. We are given that there were 1.45 x 10^(-5) moles of ascorbic acid.
Now we can use the stoichiometry of the reaction between ascorbic acid and citric acid to find the moles of citric acid in the 10.0 mL aliquot.
The balanced equation for the reaction between ascorbic acid (AA) and citric acid (CA) is:
AA + CA → 2 CO2 + 2 H2O
From the equation, we can see that the ratio between the moles of ascorbic acid and moles of citric acid is 1:1.
Therefore, the moles of citric acid in the 10.0 mL aliquot is also 1.45 x 10^(-5) moles.
To summarize, the moles of citric acid in the 10.0 mL aliquot of powdered drink mix solution is 1.45 x 10^(-5) moles.