Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 4: Problem 95

I A student is given \(0.930 \mathrm{~g}\) of an unknown acid, which can be either oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), or citric acid, \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\). To determine which acid she has, she titrates the unknown acid with \(0.615 \mathrm{M} \mathrm{NaOH}\). The equivalence point is reached when \(33.6 \mathrm{~mL}\) are added. What is the unknown acid?

Short Answer

Expert verified
Answer: The unknown acid sample is oxalic acid.

Step by step solution

01

Calculate moles of NaOH

Given the molarity and volume of NaOH used, first calculate the moles of NaOH reacted with the unknown acid. The formula to calculate moles is: moles = molarity × volume (in liters) moles of NaOH = (0.615 M) × (33.6 mL / 1000) moles of NaOH = 0.02061 mol
02

Determine the stoichiometry

Now, we need to establish the stoichiometry between the acid and NaOH. Oxalic acid is a diprotic acid (H2C2O4) and citric acid is a triprotic acid (H3C6H5O7), meaning they release 2 and 3 H+ ions, respectively. If the unknown acid is oxalic acid, 2 moles of NaOH are required to neutralize 1 mole of oxalic acid. If the unknown acid is citric acid, then 3 moles of NaOH are required to neutralize 1 mole of citric acid.
03

Calculate moles of unknown acid

Using the stoichiometry, calculate the moles of unknown acid: moles of acid_oxalic = moles of NaOH / 2 moles of acid_oxalic = 0.02061 mol / 2 moles of acid_oxalic = 0.010305 mol moles of acid_citric = moles of NaOH / 3 moles of acid_citric = 0.02061 mol / 3 moles of acid_citric = 0.00687 mol
04

Calculate molecular weight of unknown acid

Next, use the given mass of the unknown acid sample with the moles calculated in step 3 to find the molecular weight of the unknown acid for both oxalic and citric acid cases. The formula is: molecular weight = mass / moles molecular weight_oxalic = 0.930 g / 0.010305 mol molecular weight_oxalic = 90.22 g/mol molecular weight_citric = 0.930 g / 0.00687 mol molecular weight_citric = 135.37 g/mol
05

Compare with known molecular weights

Compare the calculated molecular weights with the actual molecular weights of oxalic acid (90.04 g/mol) and citric acid (192.12 g/mol). Molecular weight_oxalic = 90.22 g/mol is very close to the actual molecular weight of oxalic acid (90.04 g/mol) Molecular weight_citric = 135.37 g/mol is not close to the actual molecular weight of citric acid (192.12 g/mol) Based on the comparison, the unknown acid is oxalic acid as its calculated molecular weight is closer to the actual molecular weight of oxalic acid.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity
Molarity is a way to express the concentration of a solution. It describes how much of a substance is dissolved in a certain volume of solution. Molarity is denoted by the symbol \(M\) and is calculated using the formula:
  • \(\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}\)
In the context of a titration, knowing the molarity of a titrant, like NaOH, is essential for calculating the number of moles that react with the unknown acid. In the given exercise, the molarity of NaOH is 0.615 M, which helps determine how many moles of NaOH participate in the reaction.
Equivalence Point
The equivalence point in a titration is the moment when a stoichiometric amount of titrant has been added to react exactly with the substance in the solution being titrated. It signifies the completion of the reaction. At this point, the number of moles of the titrant is equal to the number of moles needed to completely neutralize the analyte. In this exercise, the equivalence point is reached after adding 33.6 mL of NaOH. This measurement helps to calculate the total number of moles of NaOH that reacted with the acid, thereby identifying the unknown substance.
Molecular Weight
Molecular weight is the weight of a molecule of a substance expressed in atomic mass units (u). It is crucial for converting between moles and grams. In this exercise, knowing the molecular weight of possible substances allows us to determine which acid the student has.
  • The calculated molecular weights for the potential acids - oxalic acid and citric acid - were compared with their known values.
  • Oxalic acid's calculated molecular weight was 90.22 g/mol, close to the known 90.04 g/mol.
  • Citric acid's calculated molecular weight was 135.37 g/mol, far from the known 192.12 g/mol.
This comparison indicates that the unknown acid in this exercise is oxalic acid.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass and the relation between the number of moles of reactants and products.
  • Oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4\)) is diprotic, meaning it donates 2 protons (H+ ions).
  • Citric acid (\(\text{H}_3\text{C}_6\text{H}_5\text{O}_7\)) is triprotic, donating 3 protons.
During the titration, if the acid were oxalic, each mole would react with 2 moles of NaOH. If citric, each mole would require 3 moles of NaOH. This stoichiometric relation was key in finding the correct identity of the unknown acid.
Oxalic Acid
Oxalic acid is an organic compound that functions as a diprotic acid. It is known for its use in some cleaning products due to its strength as an acid. Structurally, it consists of two carboxylic acid groups \((\text{COOH})\) linked together.In the titration experiment, oxalic acid behaves in a specific way due to its ability to donate two hydrogen ions per molecule. This means a 1:2 ratio of acid to NaOH is necessary for complete neutralization, a critical point in determining its presence as the unknown acid.
Citric Acid
Citric acid is a weak organic acid with three carboxylic acid groups \((\text{COOH})\), making it triprotic. It is commonly found in citrus fruits like lemons and limes, giving them their sour taste. In industries, citric acid is often used as a preservative or an acidity regulator.In titration, citric acid's triprotic nature requires 3 moles of a base like NaOH to completely neutralize 1 mole of citric acid. Despite being a possibility in the exercise, the calculated molecular weight didn't match the typical molecular weight of citric acid, thus ruling it out as the unknown compound.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write a balanced net ionic equation for each of the following acid-base reactions in water. (a) acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) with strontium hydroxide (b) diethylamine, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\), with sulfuric acid (c) aqueous hydrogen cyanide (HCN) with sodium hydroxide

What is the molarity of each ion present in aqueous solutions prepared by dissolving \(20.00 \mathrm{~g}\) of the following compounds in water to make 4.50 L of solution? (a) cobalt(III) chloride (b) nickel(III) sulfate (c) sodium permanganate (d) iron(II) bromide

Wlassify each of the following half-equations as oxidation or reduction and balance. (a) (acidic) \(\quad \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (b) (basic) \(\quad \mathrm{CrO}_{4}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (c) (basic) \(\quad \mathrm{PbO}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)\) (d) (acidic) \(\quad \mathrm{ClO}_{2}^{-}(a q) \longrightarrow \mathrm{ClO}^{-}(a q)\)

Decide whether a precipitate will form when the following solutions are mixed. If a precipitate forms, write a net ionic equation for the reaction. (a) potassium nitrate and magnesium sulfate (b) silver nitrate and potassium carbonate (c) ammonium carbonate and cobalt(III) chloride (d) sodium phosphate and barium hydroxide (e) barium nitrate and potassium hydroxide

Write balanced net ionic equations for the following reactions in acid solution. (a) Nitrogen oxide and hydrogen gases react to form ammonia gas and steam. (b) Hydrogen peroxide reacts with an aqueous solution of sodium hypochlorite to form oxygen and chlorine gases. (c) Zinc metal reduces the vanadyl ion \(\left(\mathrm{VO}^{2+}\right)\) to vanadium(III) ions. Zinc ions are also formed.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free