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?
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.
To find the molarity (M) of citric acid (H3A) in the lemon juice solution, we will use the equation provided:
H3A + 3NaOH → Na3A + 3H2O
First, let's determine the number of moles of NaOH used in the reaction.
Given:
Volume of NaOH solution (V1) = 28.4 ml
Molarity of NaOH (M1) = 0.100 M
Using the formula for moles (mol), we can calculate the number of moles of NaOH (n1):
n1 = M1 * V1
= 0.100 M * 0.0284 L (since 1 ml = 0.001 L)
= 0.00284 mol
From the balanced equation, we can see that one mole of citric acid reacts with three moles of NaOH. Therefore, the number of moles of citric acid (H3A) in the lemon juice solution is three times the number of moles of NaOH used in the reaction.
Number of moles of citric acid (H3A) (n2) = 3 * n1
= 3 * 0.00284 mol
= 0.00852 mol
The volume of the lemon juice sample used in the reaction is given as 25.0 ml. To find the molarity (M) of citric acid (H3A) in the lemon juice solution, we need to convert this volume to liters.
Volume of lemon juice sample (V2) = 25.0 ml
= 0.0250 L (since 1 ml = 0.001 L)
Now we can calculate the molarity (M2) of citric acid (H3A) in the lemon juice solution:
M2 = n2 / V2
= 0.00852 mol / 0.0250 L
= 0.341 M (rounded to three decimal places)
Therefore, the molarity (M) of citric acid (H3A) in the lemon juice solution is 0.341 M.
To find the %(mass/volume) of the citric acid in the lemon juice, we need to determine the mass of citric acid (H3A) in the lemon juice sample.
First, let's find the mass of NaOH used in the reaction.
Given:
Molar mass of NaOH = 40.0 g/mol
Mass of NaOH (m1) = n1 * Molar mass of NaOH
= 0.00284 mol * 40.0 g/mol
= 0.1136 g
From the balanced equation, we can see that the molar ratio between NaOH and citric acid (H3A) is 3:1. Therefore, the mass of citric acid can be calculated by dividing the mass of NaOH by 3.
Mass of citric acid (H3A) (m2) = m1 / 3
= 0.1136 g / 3
= 0.03787 g
Finally, we can calculate the %(mass/volume) of the citric acid in the lemon juice:
%(mass/volume) = (mass of citric acid / volume of lemon juice sample) * 100
= (0.03787 g / 25.0 ml) * 100
= 0.1515% (rounded to four decimal places)
Therefore, the %(mass/volume) of citric acid in the lemon juice is 0.1515%.