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Chapter 14: Problem 20

Which of the following conditions indicate an acidic solution at \(25^{\circ} \mathrm{C} ?\) a. \(\mathrm{pH}=3.04\) b. \(\left[\mathrm{H}^{+}\right]>1.0 \times 10^{-7} M\) c. \(\mathrm{pOH}=4.51\) d. \(\left[\mathrm{OH}^{-}\right]=3.21 \times 10^{-12} M\)

Short Answer

Expert verified
Conditions a (\(\mathrm{pH}=3.04\)), b (\(\left[\mathrm{H}^{+}\right]>1.0 \times 10^{-7} M\)), and d (\(\left[\mathrm{OH}^{-}\right]=3.21 \times 10^{-12} M\)) indicate an acidic solution at 25°C, while condition c (\(\mathrm{pOH}=4.51\)) does not.

Step by step solution

01

Check Condition a

For condition a, the given pH is 3.04. As pH < 7, it fulfills the criteria for an acidic solution and indicates an acidic environment.
02

Check Condition b

For condition b, the given H+ ion concentration is greater than 1.0 × 10⁻⁷ M. As the concentration of H+ ions is greater than 10⁻⁷ M, it matches the criteria for an acidic solution and indicates an acidic environment.
03

Check Condition c

For condition c, the given pOH is 4.51. To determine if this condition indicates an acidic solution, we need to calculate the corresponding pH from the pOH using the relationship: pH + pOH = 14 at 25°C. So the pH = 14 - pOH = 14 - 4.51 = 9.49. As the pH value is greater than 7, this condition does not indicate an acidic solution.
04

Check Condition d

For condition d, the given OH⁻ ion concentration is 3.21 × 10⁻¹² M. To determine if this condition indicates an acidic solution, we need to calculate the corresponding pH. We will first find the pOH using the relationship: pOH = -log[OH⁻]. Then, we will determine the pH using the relationship: pH + pOH = 14 at 25°C. pOH = -log(3.21 × 10⁻¹²) = 11.49 pH = 14 - 11.49 = 2.51 As the pH value is less than 7, this condition indicates an acidic environment. To summarize: Conditions a, b, and d indicate an acidic solution at 25°C, while condition c does not.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The prin- cipal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\)

What mass of \(\mathrm{KOH}\) is necessary to prepare \(800.0 \mathrm{~mL}\) of a solution having a \(\mathrm{pH}=11.56\) ?

Consider \(1000 . \mathrm{mL}\) of a \(1.00 \times 10^{-4} M\) solution of a certain acid HA that has a \(K_{\mathrm{a}}\) value equal to \(1.00 \times 10^{-4} .\) How much water was added or removed (by evaporation) so that a solution remains in which \(25.0 \%\) of HA is dissociated at equilibrium? Assume that HA is nonvolatile.

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