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Chapter 14: Problem 183

Will \(0.10 M\) solutions of the following salts be acidic, basic, of neutral? See Appendix 5 for \(K_{\mathrm{a}}\) values. a. ammonium bicarbonate b. sodium dihydrogen phosphate c. sodium hydrogen phosphate d. ammonium dihydrogen phosphate e. ammonium formate

Short Answer

Expert verified
a. The solution of ammonium bicarbonate will be basic. b. The solution of sodium dihydrogen phosphate will be acidic. c. The solution of sodium hydrogen phosphate will be slightly basic. d. The solution of ammonium dihydrogen phosphate will be acidic. e. The solution of ammonium formate will be basic.

Step by step solution

01

Write the chemical formula and dissociation of ammonium bicarbonate

Ammonium bicarbonate has the chemical formula \(\mathrm{NH_{4}HCO_{3}}\). When it dissolves in water, it dissociates into ammonium ions \(\mathrm{(NH_{4}^{+})}\) and bicarbonate ions \(\mathrm{(HCO_{3}^{-})}\): \[ \mathrm{NH_{4}HCO_{3} \rightarrow NH_{4}^{+} + HCO_{3}^{-}} \]
02

Determine \(K_\mathrm{a}\) and \(K_\mathrm{b}\) for each ion

Ammonium ion \(\mathrm{(NH_{4}^{+})}\) is the conjugate acid of ammonia \(\mathrm{(NH_{3})}\) and has a \(K_\mathrm{a}\) value, while bicarbonate \(\mathrm{(HCO_{3}^{-})}\) is the conjugate base of carbonic acid \(\mathrm{(H_{2}CO_{3})}\) and has a \(K_\mathrm{b}\) value. From Appendix 5 and \(K_\mathrm{w}\) formula, we get: \(K_\mathrm{a} (\mathrm{NH_{4}^{+}}) = 5.6\times10^{-10}\) \(K_\mathrm{b} (\mathrm{HCO_{3}^{-}}) = \dfrac{1.0\times10^{-14}}{4.45\times10^{-11}} = 2.25\times10^{-4}\)
03

Compare \(K_\mathrm{a}\) and \(K_\mathrm{b}\) values

Since \(K_\mathrm{a} (\mathrm{NH_{4}^{+}}) \lt K_\mathrm{b} (\mathrm{HCO_{3}^{-}})\), the solution of ammonium bicarbonate will be basic. b. Sodium dihydrogen phosphate We follow the same steps for the remaining salts:
04

Write the chemical formula and dissociation of sodium dihydrogen phosphate

Sodium dihydrogen phosphate has the chemical formula \(\mathrm{NaH_{2}PO_{4}}\). When it dissolves in water, it dissociates into sodium ions \(\mathrm{(Na^{+})}\), which are spectators, and dihydrogen phosphate ions \(\mathrm{(H_{2}PO_{4}^{-})}\): \[ \mathrm{NaH_{2}PO_{4} \rightarrow Na^{+} + H_{2}PO_{4}^{-}} \]
05

Determine \(K_\mathrm{a}\) and \(K_\mathrm{b}\) for dihydrogen phosphate ion

Dihydrogen phosphate ion \(\mathrm{(H_{2}PO_{4}^{-})}\) is the conjugate base of phosphoric acid \(\mathrm{(H_{3}PO_{4})}\) and has a \(K_\mathrm{b}\) value. From Appendix 5 and \(K_\mathrm{w}\) formula, we get: \(K_\mathrm{b} (\mathrm{H_{2}PO_{4}^{-}}) = \dfrac{1.0\times10^{-14}}{7.11\times10^{-3}} = 1.41\times10^{-12}\)
06

Compare \(K_\mathrm{a}\) and \(K_\mathrm{b}\) values

Since the \(K_\mathrm{b}\) value is very small, the solution of sodium dihydrogen phosphate will be acidic. c. Sodium hydrogen phosphate
07

Write the chemical formula and dissociation of sodium hydrogen phosphate

Sodium hydrogen phosphate has the chemical formula \(\mathrm{Na_{2}HPO_{4}}\). When it dissolves in water, it dissociates into sodium ions \(\mathrm{(Na^{+})}\), which are spectators, and hydrogen phosphate ions \(\mathrm{(HPO_{4}^{2-})}\): \[ \mathrm{Na_{2}HPO_{4} \rightarrow 2 Na^{+} + HPO_{4}^{2-}} \]
08

Determine \(K_\mathrm{a}\) and \(K_\mathrm{b}\) for hydrogen phosphate ion

Hydrogen phosphate ion \(\mathrm{(HPO_{4}^{2-})}\) is the conjugate base of dihydrogen phosphate ion \(\mathrm{(H_{2}PO_{4}^{-})}\) and has a \(K_\mathrm{b}\) value. From Appendix 5 and \(K_\mathrm{w}\) formula, we get: \(K_\mathrm{b} (\mathrm{HPO_{4}^{2-}}) = \dfrac{1.0\times10^{-14}}{6.314\times10^{-8}} = 1.58\times10^{-7}\)
09

Compare \(K_\mathrm{a}\) and \(K_\mathrm{b}\) values

Since the \(K_\mathrm{b}\) value is small but not significantly smaller than \(1\), the solution of sodium hydrogen phosphate will be close to neutral, but slightly basic. d. Ammonium dihydrogen phosphate
10

Write the chemical formula and dissociation of ammonium dihydrogen phosphate

Ammonium dihydrogen phosphate has the chemical formula \(\mathrm{(NH_{4})(H_{2}PO_{4})}\). When it dissolves in water, it dissociates into ammonium ions \(\mathrm{(NH_{4}^{+})}\) and dihydrogen phosphate ions \(\mathrm{(H_{2}PO_{4}^{-})}\): \[ \mathrm{(NH_{4})(H_{2}PO_{4}) \rightarrow NH_{4}^{+} + H_{2}PO_{4}^{-}} \]
11

Compare \(K_\mathrm{a}\) and \(K_\mathrm{b}\) values of ammonium and dihydrogen phosphate ions

We have already found the \(K_\mathrm{a}\) value for ammonium ions in part a and the \(K_\mathrm{b}\) value for dihydrogen phosphate ions in part b. Since \(K_\mathrm{a} (\mathrm{NH_{4}^{+}}) \gt K_\mathrm{b} (\mathrm{H_{2}PO_{4}^{-}})\), the solution of ammonium dihydrogen phosphate will be acidic. e. Ammonium formate
12

Write the chemical formula and dissociation of ammonium formate

Ammonium formate has the chemical formula \(\mathrm{NH_{4}CHO_{2}}\). When it dissolves in water, it dissociates into ammonium ions \(\mathrm{(NH_{4}^{+})}\) and formate ions \(\mathrm{(CHO_{2}^{-})}\): \[ \mathrm{NH_{4}CHO_{2} \rightarrow NH_{4}^{+} + CHO_{2}^{-}} \]
13

Determine \(K_\mathrm{a}\) and \(K_\mathrm{b}\) for formate ion

Formate ion \(\mathrm{(CHO_{2}^{-})}\) is the conjugate base of formic acid \(\mathrm{(HCO_{2}H)}\) and has a \(K_\mathrm{b}\) value. From Appendix 5 and \(K_\mathrm{w}\) formula, we get: \(K_\mathrm{b} (\mathrm{CHO_{2}^{-}}) = \dfrac{1.0\times10^{-14}}{1.77\times10^{-4}} = 5.65\times10^{-11}\)
14

Compare \(K_\mathrm{a}\) and \(K_\mathrm{b}\) values

Since \(K_\mathrm{a} (\mathrm{NH_{4}^{+}}) \lt K_\mathrm{b} (\mathrm{CHO_{2}^{-}})\), the solution of ammonium formate will be basic.

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Most popular questions from this chapter

Acrylic acid \(\left(\mathrm{CH}_{2}=\mathrm{CHCO}_{2} \mathrm{H}\right)\) is a precursor for many important plastics. \(K_{\mathrm{a}}\) for acrylic acid is \(5.6 \times 10^{-5}\) a. Calculate the \(\mathrm{pH}\) of a \(0.10-M\) solution of acrylic acid. b. Calculate the percent dissociation of a \(0.10-M\) solution of acrylic acid. c. Calculate the \(\mathrm{pH}\) of a \(0.050-M\) solution of sodium acrylate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

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