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Chapter 17: Problem 102

To increase the ionization of formic acid, \(\mathrm{HCOOH}(\mathrm{aq})\) which of the following should be added to the solution? (a) \(\mathrm{NaCl} ;\) (b) \(\mathrm{NaHCOO} ;(\mathrm{c}) \mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(\mathrm{NaHCO}_{3}\)

Short Answer

Expert verified
To increase the ionization of formic acid, sulphuric acid, H2SO4, should be added to the solution.

Step by step solution

01

Analyze NaCl

Sodium chloride, NaCl, is a salt which when added to the solution will increase the ionic strength of the solution. However, it will not contribute to the ionization of HCOOH, because NaCl won't interact with the acid or the base formed during the ionization.
02

Analyze NaHCOO

Sodium formate, NaHCOO, is the salt result from the ionization of formic acid. If we add NaHCOO to the solution, it will provide more formate ions, HCOO-. This will push the equilibrium towards the left, reducing the ionization of formic acid due to Le Chatelier's principle.
03

Analyze H2SO4

Sulphuric acid, H2SO4, it is a strong acid. If this is added to the solution, it can react with the formate ion, HCOO-, produced in the ionization of formic acid and convert it back to formic acid. Thus, addition of H2SO4 increases the ionization of formic acid, because the ion produced by ionization is consumed, removing it from the equilibrium and pushing the reaction towards right making more acid ionize.
04

Analyze NaHCO3

Sodium bicarbonate, NaHCO3, is a weak base. If added to the solution, it reacts with the acid but as a weak base it can't shift the equilibrium as much as a strong acid like H2SO4.

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Most popular questions from this chapter

\(\begin{array}{lll}\text { Given } & 1.00 & \mathrm{L}\end{array}\) of a solution that is \(0.100 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) and \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}\) (a) Over what pH range will this solution be an effective buffer? (b) What is the buffer capacity of the solution? That is, how many millimoles of strong acid or strong base can be added to the solution before any significant change in pH occurs?

Sodium ammonium hydrogen phosphate, \(\mathrm{NaNH}_{4}\) \(\mathrm{HPO}_{4},\) is a salt in which one of the ionizable \(\mathrm{H}\) atoms of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is replaced by \(\mathrm{Na}^{+},\) another is replaced by \(\mathrm{NH}_{4}^{+},\) and the third remains in the anion \(\mathrm{HPO}_{4}^{2-}\) Calculate the \(\mathrm{pH}\) of \(0.100 \mathrm{M} \mathrm{NaNH}_{4} \mathrm{HPO}_{4}(\mathrm{aq})\) [Hint: You can use the general method introduced on page \(720 .\) First, identify all the species that could be present and the equilibria involving these species. Then identify the two equilibrium expressions that will predominate and eliminate all the species whose concentrations are likely to be negligible. At that point, only a few algebraic manipulations are required.]

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\text { aq) requires } 31.15 \mathrm{mL}\) of \(0.2420 \mathrm{M}\) KOH for titration to the second equivalence point. What is the molarity of the \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) ?\)

The effect of adding \(0.001 \mathrm{mol} \mathrm{KOH}\) to 1.00 Lof a solution that is \(0.10 \mathrm{M} \mathrm{NH}_{3}-0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) is to (a) raise the pH very slightly; (b) lower the pH very slightly; (c) raise the pH by several units; (d) lower the pH by several units.

In the use of acid-base indicators, (a) Why is it generally sufficient to use a single indicator in an acid-base titration, but often necessary to use several indicators to establish the approximate pH of a solution? (b) Why must the quantity of indicator used in a titration be kept as small as possible?
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