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Chapter 16: Problem 114

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with all the carbonate and then "back titrating" the remaining acid with a \(\mathrm{NaOH}\) solution. (a) Write equations for these reactions. (b) In a certain experiment, \(18.68 \mathrm{~mL}\) of \(5.653 \mathrm{M} \mathrm{HCl}\) were added to a \(3.542-\mathrm{g}\) sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(12.06 \mathrm{~mL}\) of \(1.789 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

Short Answer

Expert verified
The molar mass of the carbonate is 84.42 g/mol. The metal M is Magnesium (Mg).

Step by step solution

01

Write the Reactions

Reaction of the metal carbonate with \( HCl \): \( MCO_3 + 2HCl \rightarrow MCl_2 + CO_2 + H_2O \). \nBack titration of the excess \( HCl \) with \( NaOH \): \( HCl + NaOH \rightarrow NaCl + H_2O \)
02

Calculate the moles of \( NaOH \) and \( HCl \)

We use the formula: moles = Molarity × Volume (in litters). \n Volume of \( NaOH \) used = 12.06 mL = 0.01206 L. \nMolarity of \( NaOH \) = 1.789 M. \nHence, Moles of \( NaOH \) = 1.789 M × 0.01206 L = 0.02176 moles. \nSince the ratio of \( HCl \) to \( NaOH \) in the reaction is 1:1, there were 0.02176 moles of \( HCl \) left after the reaction with \( MCO_3 \).
03

Calculate the moles of \( HCl \) reacted with \( MCO_3 \)

Volumes of \( HCl \) added = 18.68 mL = 0.01868 L. \nMolarities of \( HCl \) = 5.653 M. \nTotal moles of \( HCl \) = Molarity × Volume = 5.653 M × 0.01868 L = 0.1056 moles. \n So moles of \( HCl \) reacted with \( MCO_3 \) = Total moles of \( HCl \) – moles of \( HCl \) left = 0.1056 moles – 0.02176 moles = 0.08384 moles.
04

Calculate the moles of \( MCO_3 \) and its molar mass

For reaction between \( MCO_3 \) and \( HCl \), the stoichiometric coefficient of \( MCO_3 \) and \( HCl \) is 1:2. Hence, the moles of \( MCO_3 \) should be Half of the moles of \( HCl \) that took part in the reaction = 0.08384 moles ÷ 2 = 0.4192 moles. \nThe molar mass of \( MCO_3 \) = mass of \( MCO_3 \) ÷ moles of \( MCO_3 \) = 3.542g ÷ 0.4192 moles = 84.42 g/mol.
05

Identify M

The molar mass of \( CO_3 \) = 12.01 g/mol (for C) + 3.0 g/mol (for O) x3 = 60.01 g/mol. \nSubtracting molar mass of \( CO_3 \) from the molar mass of \( MCO_3 \) gives mass of Metal M = 84.42 g/mol - 60.01 g/mol = 24.41 g/mol. \nThe element that has a molar mass close to 24.41 g/mol is Magnesium (Mg)

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Most popular questions from this chapter

A student carried out an acid-base titration by adding \(\mathrm{NaOH}\) solution from a buret to an Erlenmeyer flask containing HCl solution and using phenolphthalein as indicator. At the equivalence point, she observed a faint reddish-pink color. However, after a few minutes, the solution gradually turned colorless. What do you suppose happened?

The solubility product of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1.2 \times 10^{-11}\) What minimum \(\mathrm{OH}^{-}\) concentration must be attained (for example, by adding \(\mathrm{NaOH}\) ) to decrease the \(\mathrm{Mg}^{2+}\) concentration in a solution of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) to less than \(1.0 \times 10^{-10} M ?\)

Describe how you would prepare a 1-L \(0.20 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COONa} / 0.20 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) buffer system by (a) mixing a solution of \(\mathrm{CH}_{3} \mathrm{COOH}\) with a solution of \(\mathrm{CH}_{3} \mathrm{COONa},\) (b) reacting a solution of \(\mathrm{CH}_{3} \mathrm{COOH}\) with a solution of \(\mathrm{NaOH},\) and (c) reacting a solution of \(\mathrm{CH}_{3}\) COONa with a solution of \(\mathrm{HCl}\).

A \(25.0-\mathrm{mL}\) solution of \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) is ti- trated with a \(0.200 \mathrm{M}\) KOH solution. Calculate the \(\mathrm{pH}\) after the following additions of the KOH solution: (a) \(0.0 \mathrm{~mL},\) (b) \(5.0 \mathrm{~mL},\) (c) \(10.0 \mathrm{~mL}\), (d) \(12.5 \mathrm{~mL}\) (e) \(15.0 \mathrm{~mL}\).

When lemon juice is squirted into tea, the color becomes lighter. In part, the color change is due to dilution, but the main reason for the change is an acid-base reaction. What is the reaction? (Hint: Tea contains "polyphenols" which are weak acids and lemon juice contains citric acid.)
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