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Chapter 16: Problem 29

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) $\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-10} \mathrm{M}(\mathbf{b})\left[\mathrm{OH}^{-}\right]=0.015 \mathrm{M} ;$ (c) a solution in which \(\left[\mathrm{H}^{+}\right]\) is 10 times greater than \(\left[\mathrm{OH}^{-}\right]\).

Short Answer

Expert verified
Short Answer: (a) For \(\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-10} \mathrm{M}\), we find \(\left[\mathrm{H}^{+}\right]=1.37 \times 10^{-6} \mathrm{M}\). The solution is acidic. (b) For \(\left[\mathrm{OH}^{-}\right]=0.015 \mathrm{M}\), we find \(\left[\mathrm{H}^{+}\right]=6.67 \times 10^{-13} \mathrm{M}\). The solution is basic. (c) For a solution in which \(\left[\mathrm{H}^{+}\right]\) is 10 times greater than \(\left[\mathrm{OH}^{-}\right]\), we find \(\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-6.5} \mathrm{M}\). The solution is acidic.

Step by step solution

01

Part (a) Calculate \(\left[\mathrm{H}^{+}\right]\) for \(\left[\mathrm{OH}^{-}\right]=7.3\times10^{-10}\) M

To find the \(\left[\mathrm{H}^{+}\right]\), we will apply the ion product of water: \[K_w = \left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]= 1.0 \times 10^{-14}\] Now, substitute the given \(\left[\mathrm{OH}^{-}\right]\) value: \[1.0 \times 10^{-14} = \left[\mathrm{H}^{+}\right]\times(7.3\times10^{-10})\] Next, solve for \(\left[\mathrm{H}^{+}\right]\): \[\left[\mathrm{H}^{+}\right] = \frac{1.0 \times 10^{-14}}{7.3\times10^{-10}}\] \[\left[\mathrm{H}^{+}\right] = 1.37\times10^{-6}\] Since \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\), the solution is acidic.
02

Part (b) Calculate \(\left[\mathrm{H}^{+}\right]\) for \(\left[\mathrm{OH}^{-}\right]=0.015\) M

Apply the ion product of water again and substitute the given \(\left[\mathrm{OH}^{-}\right]\) value: \[1.0 \times 10^{-14} = \left[\mathrm{H}^{+}\right]\times 0.015\] Solve for \(\left[\mathrm{H}^{+}\right]\): \[\left[\mathrm{H}^{+}\right] = \frac{1.0 \times 10^{-14}}{0.015}\] \[\left[\mathrm{H}^{+}\right] = 6.67\times10^{-13}\] Since \(\left[\mathrm{H}^{+}\right] < \left[\mathrm{OH}^{-}\right]\), the solution is basic.
03

Part (c) Calculate \(\left[\mathrm{H}^{+}\right]\) when it is 10 times greater than \(\left[\mathrm{OH}^{-}\right]\)

We know that \(\left[\mathrm{H}^{+}\right]\) is 10 times greater than \(\left[\mathrm{OH}^{-}\right]\), so let's denote this relationship as: \[\left[\mathrm{H}^{+}\right] = 10\left[\mathrm{OH}^{-}\right]\] Now apply the ion product of water again: \[1.0 \times 10^{-14} = (10\left[\mathrm{OH}^{-}\right])\left[\mathrm{OH}^{-}\right]\] Solve for \(\left[\mathrm{OH}^{-}\right]\): \[\left[\mathrm{OH}^{-}\right]^2 = \frac{1.0 \times 10^{-14}}{10}\] \[\left[\mathrm{OH}^{-}\right] = \sqrt{1.0 \times 10^{-15}}\] \[\left[\mathrm{OH}^{-}\right] = 1.0\times10^{-7.5}\] Now we can find the \(\left[\mathrm{H}^{+}\right]\): \[\left[\mathrm{H}^{+}\right] = 10\left[\mathrm{OH}^{-}\right] = 10\times(1.0\times10^{-7.5}) = 1.0\times10^{-6.5}\] Since \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\), the solution is acidic.

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Most popular questions from this chapter

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} ;\) (b) sulfite, \(\mathrm{SO}_{3}^{2-}\); (c) cyanide, \(\mathrm{CN}^{-}\).

Predict the stronger acid in each pair: (a) \(\mathrm{HCl}\) or HF; (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) or \(\mathrm{H}_{3} \mathrm{AsO}_{4} ;\) (c) \(\mathrm{HBrO}_{3}\) or \(\mathrm{HBrO}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) or \(\mathrm{HC}_{2} \mathrm{O}_{4} \overline{;} ;(\mathbf{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) or phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right) .\)

Calculate the pH of each of the following strong acid solutions: $(\mathbf{a}) 0.0178 \mathrm{M} \mathrm{HNO}_{3},(\mathbf{b}) 0.500 \mathrm{~g}$ of \(\mathrm{HClO}_{3}\) in \(5.00 \mathrm{~L}\) of solution, \((\mathbf{c}) 5.00 \mathrm{~mL}\) of $2.00 \mathrm{M} \mathrm{HCl}\( diluted to \)0.500 \mathrm{~L}$, (d) a mixture formed by adding \(75.0 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HBr}\).

Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{pK}_{b}\) of the butyrate ion is 9.16. (a) Calculate the \(K_{a}\) for butyric acid. (b) Calculate the pH of a \(0.075 \mathrm{M}\) solution of butyric acid. (c) Calculate the \(\mathrm{pH}\) of a \(0.075 \mathrm{M}\) solution of sodium butyrate.

Deuterium oxide \(\left(\mathrm{D}_{2} \mathrm{O},\right.\) where \(\mathrm{D}\) is deuterium, the hydrogen- 2 isotope) has an ion-product constant, \(K_{w}\), of \(8.9 \times 10^{-16}\) at \(20^{\circ} \mathrm{C}\). Calculate \(\left[\mathrm{D}^{+}\right]\) and \(\left[\mathrm{OD}^{-}\right]\) for pure (neutral) \(\mathrm{D}_{2} \mathrm{O}\) at this temperature.
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