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Chapter 14: Problem 44

Calculate the \(\left[\mathrm{H}^{+}\right]\) of each of the following solutions at \(25^{\circ} \mathrm{C}\). Identify each solution as neutral, acidic, or basic. a. \(\left[\mathrm{OH}^{-}\right]=1.5 \mathrm{M}\) b. \(\left[\mathrm{OH}^{-}\right]=3.6 \times 10^{-15} \mathrm{M}\) c. \(\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-7} M\) d. \(\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-4} M\)

Short Answer

Expert verified
The concentrations of \(\mathrm{[H^+]}\) for the given solutions are: a. \(\mathrm{[H^+]} = 6.67 \times 10^{-15} M\) (basic) b. \(\mathrm{[H^+]} = 2.78 \times 10^{0} M\) (basic) c. \(\mathrm{[H^+]} = 1.0 \times 10^{-7} M\) (neutral) d. \(\mathrm{[H^+]} = 1.37 \times 10^{-12} M\) (basic)

Step by step solution

01

Calculate \(\mathrm{[H^+]}\) for Solution a

To find the \(\mathrm{[H^+]}\) for solution a, where \(\mathrm{[OH^-]} = 1.5 \mathrm{M}\), use the equation: \([\mathrm{H^+}] = \frac{K_w}{[\mathrm{OH^-}]}\) \([\mathrm{H^+}] = \frac{1.0 \times 10^{-14}}{1.5}\) \([\mathrm{H^+}] = 6.67 \times 10^{-15}\) As \(\mathrm{[H^+]} < \mathrm{[OH^-]}\), solution a is basic.
02

Calculate \(\mathrm{[H^+]}\) for Solution b

To calculate the \(\mathrm{[H^+]}\) for solution b, where \(\mathrm{[OH^-]} = 3.6 \times 10^{-15} \mathrm{M}\), use the equation: \([\mathrm{H^+}] = \frac{K_w}{[\mathrm{OH^-}]}\) \([\mathrm{H^+}] = \frac{1.0 \times 10^{-14}}{3.6 \times 10^{-15}}\) \([\mathrm{H^+}] = 2.78 \times 10^{-15}\) As \(\mathrm{[H^+]} < \mathrm{[OH^-]}\), solution b is basic.
03

Calculate \(\mathrm{[H^+]}\) for Solution c

To calculate the \(\mathrm{[H^+]}\) for solution c, where \(\mathrm{[OH^-]} = 1.0 \times 10^{-7} \mathrm{M}\), use the equation: \([\mathrm{H^+}] = \frac{K_w}{[\mathrm{OH^-}]}\) \([\mathrm{H^+}] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-7}}\) \([\mathrm{H^+}] = 1.0 \times 10^{-7}\) As \(\mathrm{[H^+]} = \mathrm{[OH^-]}\), solution c is neutral.
04

Calculate \(\mathrm{[H^+]}\) for Solution d

To calculate the \(\mathrm{[H^+]}\) for solution d, where \(\mathrm{[OH^-]} = 7.3 \times 10^{-4} \mathrm{M}\), use the equation: \([\mathrm{H^+}] = \frac{K_w}{[\mathrm{OH^-}]}\) \([\mathrm{H^+}] = \frac{1.0 \times 10^{-14}}{7.3 \times 10^{-4}}\) \([\mathrm{H^+}] = 1.37 \times 10^{-12}\) As \(\mathrm{[H^+]} < \mathrm{[OH^-]}\), solution d is basic. To summarize, the nature of the solutions are: a - basic, b - basic, c - neutral, and d - basic.

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Most popular questions from this chapter

One mole of a weak acid HA was dissolved in \(2.0 \mathrm{~L}\) of solution. After the system had come to equilibrium, the concentration of HA was found to be \(0.45 M .\) Calculate \(K_{\mathrm{a}}\) for HA.

A \(0.15 M\) solution of a weak acid is \(3.0 \%\) dissociated. Calculate \(K_{\mathrm{a}} .\)

Making use of the assumptions we ordinarily make in calculating the pH of an aqueous solution of a weak acid, calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-6} M\) solution of hypobromous acid \(\left(\mathrm{HBrO}, K_{\mathrm{a}}=\right.\) \(\left.2 \times 10^{-9}\right) .\) What is wrong with your answer? Why is it wrong? Without trying to solve the problem, tell what has to be included to solve the problem correctly.

Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced equations for the reactions causing the solution to be acidic or basic. The relevant \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values are found in Tables \(14.2\) and \(14.3 .\) a. \(\mathrm{KCl}\) c. \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) e. \(\mathrm{NH}_{4} \mathrm{~F}\) b. \(\mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) d. \(\mathrm{KF}\) f. \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{CN}\)

Calculate the \(\mathrm{pH}\) of a \(0.20 \mathrm{MC}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) solution \(\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\).
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