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Chapter 16: Problem 30

Calculate \(\left[\mathrm{OH}^{-}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: \((\mathbf{a})\left[\mathrm{H}^{+}\right]=0.0505 M (\mathbf{b})\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-10} M ;(\mathbf{c})\) a solution in which \(\left[\mathrm{H}^{+}\right]\) is 1000 times greater than \(\left[\mathrm{OH}^{-}\right] .\)

Short Answer

Expert verified
The concentrations of OH- ions for each solution are: (a) \([\mathrm{OH}^{-}] = 1.98 \times 10^{-13}\,M\), acidic (b) \([\mathrm{OH}^{-}] = 4.0 \times 10^{-5}\,M\), basic (c) \([\mathrm{OH}^{-}] = 1.0 \times 10^{-8.5}\,M\), acidic

Step by step solution

01

Find the concentration of OH- ions for each solution

We will use the ion product of water equation to find the concentration of OH- ions for each solution. \(K_{w} = [\mathrm{H}^{+}] [\mathrm{OH}^{-}]\) For (a), (b), and (c), we will plug in the given value of [\(\mathrm{H}^{+}\)] and solve for [\(\mathrm{OH}^{-}\)].
02

Determine whether the solution is acidic, basic, or neutral

To determine the nature of each solution, we will compare the [\(\mathrm{H}^{+}\)] and [\(\mathrm{OH}^{-}\)] concentrations. If [\(\mathrm{H}^{+}\)] > [\(\mathrm{OH}^{-}\)], the solution is acidic. If [\(\mathrm{H}^{+}\)] < [\(\mathrm{OH}^{-}\)], the solution is basic. If [\(\mathrm{H}^{+}\)] = [\(\mathrm{OH}^{-}\)], the solution is neutral. Let's find the concentrations of OH- ions and the nature of each solution.
03

(a)

Given [\(\mathrm{H}^{+}\)] = 0.0505 M, we can find the [\(\mathrm{OH}^{-}\)] concentration: \(1.0 \times 10^{-14} = (0.0505) [\mathrm{OH}^{-}]\) \([\mathrm{OH}^{-}] = \frac{1.0 \times 10^{-14}}{0.0505} = 1.98 \times 10^{-13}\,M\) Since [\(\mathrm{H}^{+}\)] > [\(\mathrm{OH}^{-}\)], the solution is acidic.
04

(b)

Given [\(\mathrm{H}^{+}\)] = \(2.5 \times 10^{-10}\,M\), we can find the [\(\mathrm{OH}^{-}\)] concentration: \(1.0 \times 10^{-14} = (2.5 \times 10^{-10}) [\mathrm{OH}^{-}]\) \([\mathrm{OH}^{-}] = \frac{1.0 \times 10^{-14}}{2.5 \times 10^{-10}} = 4.0 \times 10^{-5}\,M\) Since [\(\mathrm{H}^{+}\)] < [\(\mathrm{OH}^{-}\)], the solution is basic.
05

(c)

Given that [\(\mathrm{H}^{+}\)] is 1000 times greater than [\(\mathrm{OH}^{-}\)], we can write this relation as: \([\mathrm{H}^{+}] = 1000 [\mathrm{OH}^{-}]\) Using the ion product of water: \(1.0 \times 10^{-14} = (1000 [\mathrm{OH}^{-}]) [\mathrm{OH}^{-}]\) \([\mathrm{OH}^{-}]^2 = \frac{1.0 \times 10^{-14}}{1000} = 1.0 \times 10^{-17}\) \([\mathrm{OH}^{-}] = 1.0 \times 10^{-8.5}\,M\) Since [\(\mathrm{H}^{+}\)] > [\(\mathrm{OH}^{-}\)], the solution is acidic.

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Most popular questions from this chapter

Indicate whether each of the following statements is correct or incorrect. (a) Every Bronsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{NO}_{2}^{-},(\mathbf{b}) \mathrm{PO}_{4}^{3-}\) or \(\mathrm{AsO}_{4}^{3-}\) \((\mathbf{c}) \mathrm{HCO}_{3}^{-}\) or \(\mathrm{CO}_{3}^{2-}.\)

Lactic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right)\) has one acidic hydrogen. A 0.10 \(\mathrm{M}\) solution of lactic acid has a pH of \(2.44 .\) Calculate \(K_{a} .\)

Calculate the \(\mathrm{pH}\) of each of the following solutions \((K_{a}\) and \(K_{b}\) values are given in Appendix \(\mathrm{D} ) :\) (a) 0.095\(M\) propionicacid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right),(\mathbf{b}) 0.100 M\) hydrogen chromate ion \(\left(\mathrm{HCrO}_{4}^{-}\right),(\mathbf{c}) 0.120 M\) pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right) .\)

Calculate the percent ionization of hydrazoic acid \((\mathrm{HN}_{3})\) in solutions of each of the following concentrations \((K_{a}\) is given in Appendix \(\mathrm{D} ) :(\mathbf{a}) 0.400 M (\mathbf{b}) 0.100 M,(\mathbf{c}) 0.0400 M\)
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