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Chapter 14: Problem 172

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a \(3.0 \times 10^{-7}-M\) solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Short Answer

Expert verified
The concentration of hydroxide ions, \(\left[\mathrm{OH}^{-}\right]\), in a \(3.0 \times 10^{-7}\,\mathrm{M}\) solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(6.0 \times 10^{-7}\,\mathrm{M}\).

Step by step solution

01

Write the dissociation equation for calcium hydroxide

Calcium hydroxide is a strong base that dissociates completely in water. The dissociation equation for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is given by: \[\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}\] This equation shows that for every one molecule of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that dissociates, one \(\mathrm{Ca}^{2+}\) ion and two \(\mathrm{OH}^{-}\) ions are formed.
02

Calculate the concentration of hydroxide ions using stoichiometry

Since we are given the concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) as \(3.0 \times 10^{-7}\,\mathrm{M}\), we can use the stoichiometry from the dissociation equation to find the concentration of hydroxide ions. As per the equation, the formation of hydroxide ions happens in a 1:2 ratio with respect to \(\mathrm{Ca}(\mathrm{OH})_{2}\), so we can write: \[\left[\mathrm{OH}^{-}\right] = 2 \times \left[\mathrm{Ca}(\mathrm{OH})_{2}\right]\] Now, substitute the given concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) into the equation: \[\left[\mathrm{OH}^{-}\right] = 2 \times (3.0 \times 10^{-7}\,\mathrm{M})\]
03

Calculate the final concentration of hydroxide ions

Now, simply multiply the concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) by the stoichiometric coefficient (2) to obtain the concentration of hydroxide ions: \[\left[\mathrm{OH}^{-}\right] = 6.0 \times 10^{-7}\,\mathrm{M}\] Thus, the concentration of hydroxide ions, \(\left[\mathrm{OH}^{-}\right]\), in the given solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(6.0 \times 10^{-7}\,\mathrm{M}\).

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Most popular questions from this chapter

A codeine-containing cough syrup lists codeine sulfate as a major ingredient instead of codeine. The Merck Index gives \(\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{~N}_{2} \mathrm{O}_{10} \mathrm{~S}\) as the formula for codeine sulfate. Describe the composition of codeine sulfate. (See Exercise 155.) Why is codeine sulfate used instead of codeine?

Consider \(0.10 M\) solutions of the following compounds: \(\mathrm{AlCl}_{3}\) \(\mathrm{NaCN}, \mathrm{KOH}, \mathrm{CsClO}_{4}\), and NaF. Place these solutions in order of increasing \(\mathrm{pH}\).

The \(K_{\mathrm{b}}\) values for ammonia and methylamine are \(1.8 \times 10^{-5}\) and \(4.4 \times 10^{-4}\), respectively. Which is the stronger acid, \(\mathrm{NH}_{4}{ }^{+}\) or \(\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\) ?

A \(0.20-M\) sodium chlorobenzoate \(\left(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution has a pH of \(8.65 .\) Calculate the \(\mathrm{pH}\) of a \(0.20-M\) chlorobenzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution.

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a derivative of morphine that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but is now available only by prescription because of its addictive properties. If the \(\mathrm{pH}\) of a \(1.7 \times 10^{-3}-M\) solution of codeine is \(9.59\), calculate \(K_{\mathrm{b}}\).
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