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Chapter 16: Problem 84

Show that when \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) is reduced to half its original value, the \(\mathrm{pH}\) of a solution increases by 0.30 unit, regardless ofthe initial \(p H .\) Is it also true that when any solution is diluted to half its original concentration, the pH increases by 0.30 unit? Explain.

Short Answer

Expert verified
Yes, when \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration of a solution is reduced to half, the pH increases by 0.30 unit regardless of the initial value. However, halving the concentration of any solution does not always result in a 0.30 unit increase in pH. This principle only applies to solutions where the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) or \(\mathrm{OH}^{-}\) are altered.

Step by step solution

01

Relating change in Hydronium ion concentration to pH

The pH of a solution is defined as the negative of the base 10 logarithm of the hydronium ion (\(\mathrm{H}_{3} \mathrm{O}^{+}\)) concentration. The equation for pH can be written as: \[ \mathrm{pH} = - \log[\mathrm{H}_{3} \mathrm{O}^{+}]\]Now, if the concentration of \( \mathrm{H}_{3} \mathrm{O}^{+}\) is halved, for example from an initial concentration C to C/2, the change in pH, ΔpH, can be evaluated from:\[ Δ\mathrm{pH} = -\log ([\mathrm{H}_{3} \mathrm{O}^{+}]_{final}) + \log ([\mathrm{H}_{3} \mathrm{O}^{+}]_{initial}) \] \[ Δ\mathrm{pH} = -\log (C/2) + \log (C) \]
02

Calculation of pH change

Rewrite the previous equation using the properties of logarithm (log(X/Y) = log(X) - log(Y)):\[ Δ\mathrm{pH} = \log (2) \] Applying the log base 10 value of 2, which is approximated to 0.30, we obtain:\[ Δ\mathrm{pH} = 0.30 \]This shows that, when the hydronium ion concentration is halved, the pH of a solution increases by 0.30 unit, regardless of the initial pH, meaning, it's independent of what the starting concentration (C) is.
03

Applicability of similar change for any solution concentration

The second part of the problem questions whether the same rules apply when any solution is diluted to half of its original concentration. It is important to remember that the pH is only determined by the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) (in acidic solutions) or \(\mathrm{OH}^{-}\) (base in basic solutions). These are the only ions that contribute to pH or pOH, respectively. Hence, if a solution of a substance that does not contribute to hydrogen or hydroxide ions is diluted, the pH will remain the same. Therefore, it is not universally true that diluting any solution to half its initial concentration will result in an increase in pH by 0.30 units.

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Most popular questions from this chapter

The pH of saturated \(\operatorname{Sr}(\text { OH })_{2}(\text { aq })\) is found to be 13.12 A \(10.0 \mathrm{mL}\) sample of saturated \(\operatorname{Sr}(\mathrm{OH})_{2}(\mathrm{aq})\) is diluted to \(250.0 \mathrm{mL}\) in a volumetric flask. A \(10.0 \mathrm{mL}\) sample of the diluted \(\operatorname{Sr}(\mathrm{OH})_{2}(\mathrm{aq})\) is transferred to a beaker, and some water is added. The resulting solution requires \(25.1 \mathrm{mL}\) of a \(\mathrm{HCl}\) solution for its titration. What is the molarity of this HCl solution?

What mass of benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), would you dissolve in \(350.0 \mathrm{mL}\) of water to produce a solution with a \(\mathrm{pH}=2.85 ?\) $$\begin{aligned} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} & \\ K_{\mathrm{a}}=6.3 \times 10^{-5} \end{aligned}$$

What is the (a) degree of ionization and (b) percent ionization of propionic acid in a solution that is \(0.45 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) $$\begin{aligned} &\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\\\ &&\mathrm{p} K_{\mathrm{a}}=4.89 \end{aligned}$$

For the ionization of phenylacetic acid, $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \\\ K_{\mathrm{a}}=4.9 \times 10^{-5} \end{array}$$ (a) What is \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\right]\) in \(0.186 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) (b) What is the \(\mathrm{pH}\) of \(0.121 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\)

Explain why \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in a strong acid solution doubles as the total acid concentration doubles, whereas in a weak acid solution, \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) increases only by about a factor of \(\sqrt{2}\)
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