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Chapter 16: Problem 31

At the boiling point of water $\left(100^{\circ} \mathrm{C}\right), K_{w}=5.6 \times 10^{-13} .\( Calculate \)\left[\mathrm{H}^{+}\right]$ and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Short Answer

Expert verified
In a neutral solution at the boiling point of water \(\left(100^{\circ} \mathrm{C}\right)\), the concentration of H+ and OH- ions are both approximately \(7.48 \times 10^{-7}\ \mathrm{M}\).

Step by step solution

01

Write down the ion product of water equation

The ion product of water equation relates the concentrations of H+ and OH- ions as follows: \[K_w = [\mathrm{H}^+][\mathrm{OH}^-]\] Given that \(K_w = 5.6 \times 10^{-13}\) for the boiling point of water.
02

Determine the relationship between H+ and OH- ions for a neutral solution

In a neutral solution, the concentrations of H+ and OH- ions are equal. Thus, we can represent their concentrations using a single variable, x. \[[\mathrm{H}^+] = [\mathrm{OH}^-] = x\]
03

Substitute the value of x in the ion product of water equation

Now, substitute the values of H+ and OH- ions with x in the ion product of water equation: \[K_w = x^2\] As we know that \(K_w = 5.6 \times 10^{-13}\), substitute this value in the equation: \(5.6 \times 10^{-13} = x^2\)
04

Solve the equation for x, which represents the concentrations of H+ and OH- ions

To find the value of x, simply take the square root of both sides: \(x = \sqrt{5.6 \times 10^{-13}}\) By solving this equation, we find the value of x: \(x \approx 7.48 \times 10^{-7}\)
05

Write down the result for concentrations of H+ and OH- ions

Since x represents the concentrations of H+ and OH- ions in the neutral solution at the boiling point of water: \[[\mathrm{H}^+] = [\mathrm{OH}^-] = 7.48 \times 10^{-7} \ \mathrm{M}\] Thus, the concentrations of H+ and OH- ions in a neutral solution at the boiling point of water are both approximately \(7.48 \times 10^{-7}\ \mathrm{M}\).

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(8.3 \times 10^{-4} \mathrm{MHCl},(\mathbf{b}) 1.20 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) in \(500 \mathrm{~mL}\) of solution, $(\mathbf{c}) 2.0 \mathrm{~mL}\( of \)0.250 \mathrm{M} \mathrm{HClO}_{4}\( diluted to \)40.0 \mathrm{~mL}\(, (d) a solution formed by mixing \)25.0 \mathrm{~mL}\( of \)0.100 \mathrm{M} \mathrm{HBr}\( with \)25.0 \mathrm{~mL}\( of \)0.200 \mathrm{M} \mathrm{HCl}$.

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

If a neutral solution of water, with \(\mathrm{pH}=7.00,\) is cooled to \(10^{\circ} \mathrm{C},\) the pH rises to \(7.27 .\) Which of the following three statements is correct for the cooled water: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right]\) (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right], \mathrm{or}\) (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)

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 \(K_{a}\)

What is the pH of a solution that is \(1.2 \times 10^{-8} \mathrm{M}\) in \(\mathrm{KOH}\) ? Does your answer make sense? What assumption do we normally make that is not valid in this case?
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