Explanation The given, Molar mass of citric acid = 192 .1 g/mol , Molar mass of citric acid = 150 .1 g/mol . 0 .956 g sample requires 29 .1 mL of 0 .513 M NaOH reaching the equivalence point. Given equation is shown below, C 6 H 8 O 7 (aq) + 3 NaOH(aq) → Na 3 C 6 H 5 O 7 (aq) + 3 H 2 O(l) Citric acid C 4 H 6 O 6 (aq) + 2 NaOH(aq) → Na 2 C 4 H 4 O 6 (aq) + 2 H 2 O(l) Tartaric acid Molar mass of Cl 2 is 35 .453 g/mol . Molar mass of FeCl 3 is 162 .204 g/mol Number of moles is calculated by using following formula, No .of Moles = Molarity × Volume Given, Molarity = 0 .5134 M , Volume = 29 .1 ml or 0 .0291 L . Number of moles of NaOH is calculated as follows, No .of Moles = Molarity × Volume No .of Moles NaOH = 0 .5134 × 0 .0291 No .of Moles NaOH = 0 .01493 Given balanced equation is shown below, C 6 H 8 O 7 (aq) + 3 NaOH(aq) → Na 3 C 6 H 5 O 7 (aq) + 3 H 2 O(l) Citric acid According to the balanced equation three moles of sodium hydroxide reacts with one mole of citric acid gives one mole of sodium citrate

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ISBN: 9781285199047

Author: John W. Moore, Conrad L. Stanitski

Publisher: Cengage Learning

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Chapter 3, Problem 112QRT

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When 0.956 g sample requires 29.1 mL of 0.513 M NaOH reaching the equivalence point. The unknown acid has to be identified.

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Chapter 3, Problem 111QRT

Chapter 3, Problem 113QRT

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When 0 .956 g sample requires 29 .1 mL of 0 .513 M NaOH reaching the equivalence point. The unknown acid has to be identified.

Solution Summary: The author explains that the unknown acid has to be identified. The given, Molar mass of citric acid = 192 g/mol.

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Interpretation:

When 0.956 g sample requires 29.1 mL of 0.513 M NaOH reaching the equivalence point. The unknown acid has to be identified.

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Chapter 3 Solutions

When 0 .956 g sample requires 29 .1 mL of 0 .513 M NaOH reaching the equivalence point. The unknown acid has to be identified.

Solution Summary: The author explains that the unknown acid has to be identified. The given, Molar mass of citric acid = 192 g/mol.

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Interpretation: When 0.956 g sample requires 29.1 mL of 0.513 M NaOH reaching the equivalence point. The unknown acid has to be identified.

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Chapter 3 Solutions

Interpretation:

When 0.956 g sample requires 29.1 mL of 0.513 M NaOH reaching the equivalence point. The unknown acid has to be identified.