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Chapter 14: Problem 164

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 the \(3.0 \times 10^{-7} M\) solution of Calcium Hydroxide is \(6.0 \times 10^{-7}\,\mathrm{M}\).

Step by step solution

01

STEP 1: Write the dissociation equation

When Calcium Hydroxide \(\mathrm{Ca}(\mathrm{OH})_{2}\) dissolves in water, it dissociates into its individual ions: \[\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^-\]
02

STEP 2: Set up the stoichiometry

According to the dissociation equation, the stoichiometry of the reaction states that for each mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that dissociates, 2 moles of \(\mathrm{OH}^-\) ions are produced. Let's use this ratio to find the concentration of \(\mathrm{OH}^-\) ions.
03

STEP 3: Calculate the concentration of \(\mathrm{OH}^-\) ions

We know the concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) (\(3.0 \times 10^{-7} M\)). Since 1 mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) produces 2 moles of \(\mathrm{OH^-}\) ions, we'll simply multiply the given concentration by 2 to get the concentration of \(\mathrm{OH}^-\) ions: \[\left[\mathrm{OH}^{-}\right] = 2 \times \left[\mathrm{Ca}(\mathrm{OH})_{2}\right] = 2 \times (3.0 \times 10^{-7}\,\mathrm{M})\]
04

STEP 4: Solve for the concentration of \(\mathrm{OH}^-\) ions

By multiplying the given concentration by 2, we can find the concentration of \(\mathrm{OH}^-\) ions in the solution: \[\left[\mathrm{OH}^{-}\right] = 2 \times (3.0 \times 10^{-7}\,\mathrm{M}) = 6.0 \times 10^{-7}\,\mathrm{M}\] Thus, the concentration of hydroxide ions, \(\left[\mathrm{OH}^{-}\right]\), in the \(3.0 \times 10^{-7} M\) solution of Calcium Hydroxide is \(6.0 \times 10^{-7}\,\mathrm{M}\).

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Most popular questions from this chapter

What are the major species present in \(0.250 \mathrm{M}\) solutions of each of the following acids? Calculate the \(\mathrm{pH}\) of each of these solutions. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

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 \(K_{\mathrm{b}}\).

A solution is prepared by dissolving \(0.56 \mathrm{~g}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{~L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) of this solution.

Using your results from Exercise 129, place the species in each of the following groups in order of increasing base strength. a. \(\mathrm{OH}^{-}, \mathrm{SH}^{-}, \mathrm{SeH}^{-}\) b. \(\mathrm{NH}_{3}, \mathrm{PH}_{3}\) c. \(\mathrm{NH}_{3}, \mathrm{HONH}_{2}\)

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=\) \(378.85 \mathrm{~g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0 \mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).
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