Question

a 1.00-g solid sample containing a mixture of table salt (nacl) and citric acid (h3c6h5o7, a triprotic acid) is dissolved in 15 ml of water. titration of the acid solution requires 11.80 ml of 0.4327 m naoh solution to reach the endpoint. calculate the mass percent h3c6h5o7 in the solid mixture.

Answer 1

To calculate the mass percent of citric acid in the mixture, the moles of NaOH used in the titration are found and then used to determine the moles of citric acid. Then, the mass and mass percent of citric acid are calculated. The final mass percent of H3C6H5O7 in the mixture is 32.7%.

Step-by-step explanation:

To calculate the mass percent of citric acid (H3C6H5O7) in the solid mixture, you need to determine how many moles of NaOH react with the citric acid during titration. Since citric acid is a triprotic acid, each mole can react with up to three moles of NaOH.

First, calculate the moles of NaOH used:

Moles of NaOH = Volume (L) × Molarity (M)

= 0.01180 L × 0.4327 M

= 0.00510636 mol NaOH

Since it requires three moles of NaOH to neutralize one mole of citric acid, you divide the moles of NaOH by 3 to get the moles of citric acid:

Moles of citric acid = Moles of NaOH ÷ 3

= 0.00510636 mol ÷ 3

= 0.00170212 mol

Next, calculate the mass of citric acid:

Mass of citric acid = Moles × Molar Mass

= 0.00170212 mol × 192.124 g/mol

= 0.327 g (rounded to three significant figures)

Finally, calculate the mass percent of citric acid:

Mass percent = (Mass of citric acid ÷ Mass of sample) × 100%

= (0.327 g ÷ 1.00 g) × 100%

= 32.7%

Therefore, the mass percent of H3C6H5O7 in the mixture is 32.7%.