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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.

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Most popular questions from this chapter

A sample of limestone weighing \(1.005 \mathrm{~g}\) is dissolved in \(75.00 \mathrm{~mL}\) of \(0.2500 \mathrm{M}\) hydrochloric acid. The following reaction occurs: $$ \mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$ It is found that \(19.26 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NaOH}\) is required to titrate the excess \(\mathrm{HCl}\) left after reaction with the limestone. What is the mass percent of \(\mathrm{CaCO}_{3}\) in the limestone?

Aluminum ions react with carbonate ions to form an insoluble compound, aluminum carbonate. (a) Write the net ionic equation for this reaction. (b) What is the molarity of a solution of aluminum chloride if \(30.0 \mathrm{~mL}\) is required to react with \(35.5 \mathrm{~mL}\) of \(0.137 \mathrm{M}\) sodium carbonate? (c) How many grams of aluminum carbonate are formed in (b)?

What volume of \(0.4163 M\) barium chloride will react completely with (a) \(12.45 \mathrm{~mL}\) of \(1.732 \mathrm{M}\) sulfuric acid? (b) \(15.00 \mathrm{~g}\) of ammonium phosphate? (c) \(35.15 \mathrm{~mL}\) of \(1.28 \mathrm{M}\) potassium carbonate?

A reagent bottle is labeled \(0.450 \mathrm{M} \mathrm{K}_{2} \mathrm{CO}_{3}\). (a) How many moles of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) are present in \(45.6 \mathrm{~mL}\) of this solution? (b) How many milliliters of this solution are required to furnish \(0.800 \mathrm{~mol}\) of \(\mathrm{K}_{2} \mathrm{CO}_{3} ?\) (c) Assuming no volume change, how many grams of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) do you need to add to \(2.00 \mathrm{~L}\) of this solution to obtain a \(1.000 \mathrm{M}\) solution of \(\mathrm{K}_{2} \mathrm{CO}_{3} ?\) (d) If \(50.0 \mathrm{~mL}\) of this solution is added to enough water to make \(125 \mathrm{~mL}\) of solution, what is the molarity of the diluted solution?

The average adult has about \(16 \mathrm{~g}\) of sodium ions in her blood. Assuming a total blood volume of \(5.0 \mathrm{~L}\), what is the molarity of \(\mathrm{Na}^{+}\) ions in blood?
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