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Chapter 15: Problem 114

The ion product of \(\mathrm{D}_{2} \mathrm{O}\) is \(1.35 \times 10^{-15}\) at \(25^{\circ} \mathrm{C}\). (a) Calculate \(\mathrm{pD}\) where \(\mathrm{pD}=-\log \left[\mathrm{D}^{+}\right]\). (b) For what values of \(\mathrm{pD}\) will a solution be acidic in \(\mathrm{D}_{2} \mathrm{O} ?\) (c) Derive a relation between \(\mathrm{pD}\) and \(\mathrm{pOD}\).

Short Answer

Expert verified
The pD of the solution is 7.94. A solution of \( D_{2}O \) will be acidic for pD values less than 7. The relationship between pD and pOD for \( D_{2}O \) is \( \mathrm{pD}+ \mathrm{pOD} = 14.87 \).

Step by step solution

01

Calculate [D+] from ion product of \(D_{2}O\)

The ion product of water is given as \(D^{+}[OD^{-}]\) which is equal to \(1.35 \times 10^{-15}\). As \(D_{2}O\) ionizes symmetrically, we can assume [D+] = [OD-]. Solving the equation \(D^{+}[OD^{-}] = 1.35 \times 10^{-15}\) to find [D+] gives [D+]=\[\sqrt{1.35 \times \(10^{-15}\)}] = 1.16 \times \(10^{-8} \) M \]
02

Calculate pD

pD measures the acidity of a deuterium based solution and is given as \(\mathrm{pD}=-\log\left[\mathrm{D}^{+}\right]\). Substituting the value from step 1, \(\mathrm{pD}=-\log\left[1.16 \times \(10^{-8}\)\right] = 7.94 \)
03

Determine the values of pD for acidic solution

A solution is acidic if the concentration of the \( D^{+} \) ions are greater than that of \( OD^{-} \) ions. In terms of pD, this means that the solution would be acidic if the pD is less than pOD, which is numerically equivalent to 7 for pure water at \( 25^{\circ}C \). Therefore, for a solution to be acidic in \( D_{2}O \), the pD must be less than 7.
04

Derive Relationship between pD and pOD

For water and its isotopes including \( D_{2}O \), the product of the concentrations of \( D^{+} \) ions and \( OD^{-} \) ions is constant. Hence, taking the negative logarithm of both sides of the ion product equation gives \(-\log[D^{+}] -\log[OD^{-}] = \log(1.35 x 10^{-15}) \). This can be rewritten as \( \mathrm{pD}+ \mathrm{pOD} = \mathrm{pKw} = 14.87 \), where pKw is the ion product for \( D_{2}O \) at \( 25^{\circ}C \).

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Most popular questions from this chapter

A \(0.400 M\) formic acid (HCOOH) solution freezes at \(-0.758^{\circ} \mathrm{C}\). Calculate the \(K_{\mathrm{a}}\) of the acid at that temperature. (Hint: Assume that molarity is equal to molality. Carry your calculations to three significant figures and round off to two for \(K_{\mathrm{a}}\).)

HA and \(\mathrm{HB}\) are both weak acids although \(\mathrm{HB}\) is the stronger of the two. Will it take a larger volume of a \(\begin{array}{llll}0.10 & M & \text { NaOH } & \text { solution to } & \text { neutralize } & 50.0 & \text { mL }\end{array}\) of \(0.10 M\) HB than would be needed to neutralize \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) HA?

List the factors on which the \(K_{\mathrm{a}}\) of a weak acid depends.

\(\mathrm{Zn}(\mathrm{OH})_{2}\) is an amphoteric hydroxide. Write balanced ionic equations to show its reaction with (a) \(\mathrm{HCl},\) (b) \(\mathrm{NaOH}\) [the product is \(\left.\mathrm{Zn}(\mathrm{OH})_{4}^{2-}\right]\).

Calcium hypochlorite \(\left[\mathrm{Ca}(\mathrm{OCl})_{2}\right]\) is used as a disinfectant for swimming pools. When dissolved in water it produces hypochlorous acid: $$ \begin{aligned} \mathrm{Ca}(\mathrm{OCl})_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) & \rightleftharpoons \\ & 2 \mathrm{HClO}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \end{aligned} $$ which ionizes as follows: \(\mathrm{HClO}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{ClO}^{-}(a q)\) $$ K_{\mathrm{a}}=3.0 \times 10^{-8} $$ As strong oxidizing agents, both \(\mathrm{HClO}\) and \(\mathrm{ClO}^{-}\) can kill bacteria by destroying their cellular components. However, too high a HClO concentration is irritating to the eyes of swimmers and too high a concentration of \(\mathrm{ClO}^{-}\) will cause the ions to decompose in sunlight. The recommended pH for pool water is \(7.8 .\) Calculate the percent of these species present at this pH.
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