Titration of citric acid in orange juice with sodium hydroxide

Asked by admin @ 23/11/2021 in Chemistry viewed by 288 People

Consider a different analyte for this exercise. Citric acid is found in many fruits and fruit juices. Sodium hydroxide (NaOH) is the titrant and citric acid (H3C6H5O7) the analyte according to the following balanced chemical equation. H3C6H5O7 + 3 OH − → C6H5O73− + 3 H2O

(a) What is the stoichiometry of H3C6H5O7 to OH −?
(b) Complete the following table for this titration. Data Table P2: Titration of citric acid in orange juice with sodium hydroxide.
concentration of OH − 0.0769 M
volume orange juice 8.69 mL
mass orange juice 8.83 g
volume of OH − solution 24.61 mL
Questions:
mmol of OH −: ?
mmol of H3C6H5O7: ?
mass of H3C6H5O7: ?
mass % of H3C6H5O7 in orange juice: ?
molarity of H3C6H5O7 in orange juice: ?

Answered by admin @ 23/11/2021

Answer:

(a) 1:3

(b)

mmol of OH⁻ = 1.893 mmol

mmol of H₃C₆H₅O₇: 0.6310 mmol

mass of H₃C₆H₅O₇: 0.1212 g

mass % of H₃C₆H₅O₇ in orange juice: 1.37%

molarity of H₃C₆H₅O₇ in orange juice: 0.0726 M

Explanation:

Let's consider the following balanced chemical equation.

H₃C₆H₅O₇ + 3 OH⁻ → C₆H₅O₇³⁻ + 3 H₂O

(a) What is the stoichiometry of H₃C₆H₅O₇ to OH⁻?

The molar ratio of H₃C₆H₅O₇ to OH⁻ is 1:3

(b)

mmol of OH⁻: ?

$24.61mL.\frac{0.0769mol}{L} =1.893mmol$

mmol of H₃C₆H₅O₇: ?

$1.893mmolOH^{-} .\frac{1molH_{3}C_{6}H_{5}O_{7}}{3molOH^{-} } =0.6310mmolH_{3}C_{6}H_{5}O_{7}$

mass of H₃C₆H₅O₇: ?

The molar mass of the citric acid is 192.1 g/mol.

$0.6310 \times 10^{-3} molH_{3}C_{6}H_{5}O_{7}.\frac{192.1gH_{3}C_{6}H_{5}O_{7}}{1molH_{3}C_{6}H_{5}O_{7}} =0.1212gH_{3}C_{6}H_{5}O_{7}$

mass % of H₃C₆H₅O₇ in orange juice: ?

$\frac{0.1212g}{8.83g} \times 100\%=1.37\%$

molarity of H₃C₆H₅O₇ in orange juice: ?

$M=\frac{0.6310 \times 10^{-3} molH_{3}C_{6}H_{5}O_{7}}{8.69 \times 10^{-3} L} =0.0726M$

Titration of citric acid in orange juice with sodium hydroxide

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