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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: (a) \(\left[\mathrm{H}^{+}\right]=0.0505 \mathrm{M} ;\) (b) \(\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-10} \mathrm{M} ;(\mathrm{c})\) a solution in which \(\left[\mathrm{H}^{+}\right]\) is 1000 times greater than \(\left[\mathrm{OH}^{-}\right] .\)

Short Answer

Expert verified
(a) For [H⁺] = 0.0505 M, [OH⁻] = 1.98 × 10⁻¹³ M and the solution is acidic. (b) For [H⁺] = 2.5 × 10⁻¹⁰ M, [OH⁻] = 4.0 × 10⁻⁵ M and the solution is basic. (c) In a solution where [H⁺] is 1000 times greater than [OH⁻], [OH⁻] = 3.16 × 10⁻⁷ M and the solution is acidic.

Step by step solution

01

Calculate [OH⁻] for part (a)

Given [H⁺] = 0.0505 M, and we know the ion product of water (Kw) is \(1.0 \times 10^{-14}\). Using the Kw equation, we find [OH⁻]: \[1.0 \times 10^{-14} = (0.0505)([OH^-]) \] \[[OH^-] = \frac{1.0 \times 10^{-14}}{0.0505} = 1.98 \times 10^{-13} M\]
02

Determine the nature of solution (a)

Since [OH⁻] < [H⁺], the solution is acidic. #Part (b)#
03

Calculate [OH⁻] for part (b)

Given [H⁺] = 2.5 × 10⁻¹⁰ M Using the Kw equation, we find [OH⁻]: \[1.0 \times 10^{-14} = (2.5 \times 10^{-10})([OH^-]) \] \[[OH^-] = \frac{1.0 \times 10^{-14}}{2.5 \times 10^{-10}} = 4.0 \times 10^{-5} M\]
04

Determine the nature of solution (b)

Since [OH⁻] > [H⁺], the solution is basic. #Part (c)#
05

Setting up an equation to solve for [OH⁻] in part (c)

Let [H⁺] be x, then the given condition is: [H⁺] = 1000 × [OH⁻]. So, x = 1000 × [OH⁻], and we know Kw = [H⁺][OH⁻].
06

Calculate [OH⁻] for part (c)

Using the given condition and Kw equation: \[1.0 \times 10^{-14} = (1000[OH^-])([OH^-])\] \[[OH^-]^2 = \frac{1 \times 10^{-14}}{1000}\] \[[OH^-] = \sqrt{\frac{1 \times 10^{-14}}{1000}} = 3.16 \times 10^{-7}\]
07

Determine the nature of solution (c)

Since [OH⁻] < [H⁺], the solution is acidic.

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Most popular questions from this chapter

Predict whether aqueous solutions of the following substances are acidic, basic, or neutral: (a) \(\mathrm{AlCl}_{3}\), (b) \(\mathrm{NaBr}\), (c) \(\mathrm{NaClO},(\mathrm{d})\) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}\right] \mathrm{NO}_{3},(\mathrm{e}) \mathrm{Na}_{2} \mathrm{SO}_{3}\)

What is the \(\mathrm{pH}\) of a solution that is \(2.5 \times 10^{-9}\) in \(\mathrm{NaOH}\) ? Does your answer make sense? What assumption do we normally make that is not valid in this case?

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Sr}\left(\mathrm{OH}_{2}\right),\) (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, (c) \(1.00 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to \(2.00 \mathrm{~L},\) (d) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \quad M \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\)

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} \cdot\) Calculate \(\left[\mathrm{H}^{+}\right]\) \(\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature. $$ \begin{array}{llll} \hline \text { pH } & \text { pOH } & \text { [H }^{+} \text {] } & \text { [OH }^{-} \text {] } & \text { Acidic or basic? } \\ \hline 5.25 & & & \\ & 2.02 & & \\ & & 4.4 \times 10^{-10} M & & \\ & & & 8.5 \times 10^{-2} M & \\ \hline \end{array} $$

Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: (a) \(\mathrm{CH}_{3} \mathrm{COO}^{-},\) (b) \(\mathrm{HCO}_{3}^{-}, (\mathrm{c}) \mathrm{O}^{2-}, (\mathrm{d}) \mathrm{Cl}^{-} ,(\mathrm{e}) \mathrm{NH}_{3}\)
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