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Chapter 17: Problem 3

Calculate the change in pH that results from adding (a) \(0.100 \mathrm{mol} \mathrm{NaNO}_{2}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{2}(\mathrm{aq})\) (b) \(0.100 \mathrm{mol} \mathrm{NaNO}_{3}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}(\mathrm{aq})\) Why are the changes not the same? Explain.

Short Answer

Expert verified
The changes in pH are not the same because the addition of NaNO_2 to HNO_2 introduces a common ion that affects the equilibrium of the weak acid, thereby lowering the concentration of H+ ions and increasing pH. However, the addition of NaNO_3 to HNO_3 does not affect the pH as NO₃- ions do not react with water.

Step by step solution

01

Identify the Chemical Reactions

For the addition of NaNO_2 to HNO_2, the weak acid dissociation reaction can be written as: HNO_2(aq) ⇌ H+(aq) + NO_2-(aq). For the addition of NaNO_3 to HNO_3, it is important to remember that HNO_3 is a strong acid and completely dissociates into ions as HNO_3 → H+ + NO₃-. Adding NaNO₃ will result in the release of NO₃- ions, which do not affect the pH.
02

Calculate for Part (a)

By using the formula for the acid dissociation constant (Ka) which is [H+][NO_2⁻]/[HNO_2], and the concentrations of each species from the reaction with NaNO_2, the ionization of HNO_2 can be calculated. Adding NaNO_2 to HNO_2 introduces common ion NO_2-, and causes a shift in equilibrium left because of Le Chatelier's principle. This in turn decreases [H+] and as a result the pH increases.
03

Calculate for Part (b)

NO₃- ions do not react with water. Therefore, the addition of NaNO₃ does not cause any changes to the pH level of the HNO_3 solution.
04

Compare the Results

The pH change from adding NaNO_2 to HNO_2 will be different to that of adding NaNO₃ to HNO_3 because of the difference in the reaction of the added species with water. The former adds a common ion which suppress the ionization of HNO_2, thereby lowering [H+] and increasing the pH value. The latter, however, does not affect the pH as NO₃- ions do not react with water.

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Most popular questions from this chapter

To convert \(\mathrm{NH}_{4}^{+}(\text {aq })\) to \(\mathrm{NH}_{3}(\mathrm{aq}),\) (a) add \(\mathrm{H}_{3} \mathrm{O}^{+}\) (b) raise the \(\mathrm{pH} ;\) (c) add \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (d) add \(\mathrm{NaCl}\).

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?

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.]

In the titration of \(20.00 \mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH},\) calculate the number of milliliters of \(0.200 \mathrm{M} \mathrm{HCl}\) that must be added to reach a pH of (a) \(12.55,\) (b) \(10.80,\) (c) 4.25

The neutralization of \(\mathrm{NaOH}\) by \(\mathrm{HCl}\) is represented in equation (1), and the neutralization of \(\mathrm{NH}_{3}\) by HCl in equation (2). 1. \(\mathrm{OH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O} \quad K=?\) 2\. \(\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \quad K=?\) (a) Determine the equilibrium constant \(K\) for each reaction. (b) Explain why each neutralization reaction can be considered to go to completion.
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