Asked by Nam Le
Suppose a student used 28.4 ml of 0.100 M NaOH solution to neutralize the 25.0 ml of the lemon juice sample which contains citric acid (H3A)in it.
H3A + 3NaOH -----> Na3A + 3H2O
Molar mass of citric acid = 192.2 g/mole
a)What is the molarity(M)of citric acid(H3A)in lemon juice solution?
b)What is the %(mass/volume)of the citric acid in the lemon juice?
H3A + 3NaOH -----> Na3A + 3H2O
Molar mass of citric acid = 192.2 g/mole
a)What is the molarity(M)of citric acid(H3A)in lemon juice solution?
b)What is the %(mass/volume)of the citric acid in the lemon juice?
Answers
Answered by
DrBob222
mols NaOH = M x L = ?
Convert mols NaOH to mols H3A using the coefficients in the balanced equation.
Then M H3A = mols H3A/L H3A.
for b. M = mols/L solution.
Convert mols to grams. g = mols H3A x molar mass H3A to give you grams/L. You want g/100 so divide grams by 10 to give you g/100 mL solution which is % H3A.
Convert mols NaOH to mols H3A using the coefficients in the balanced equation.
Then M H3A = mols H3A/L H3A.
for b. M = mols/L solution.
Convert mols to grams. g = mols H3A x molar mass H3A to give you grams/L. You want g/100 so divide grams by 10 to give you g/100 mL solution which is % H3A.
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