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Chapter 14: Problem 178

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 \(\mathrm{OH}^{-}\) ions in the \(3.0 \times 10^{-7}\ \mathrm{M}\) solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is 6.0 x \(10^{-7}\ \mathrm{M}\).

Step by step solution

01

Write the balanced chemical equation for the reaction

The dissociation of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water can be represented by the following equation: \[\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}\] This equation tells us that for each mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that dissociates, two moles of \(\mathrm{OH}^{-}\) ions are produced.
02

Calculate the moles of \(\mathrm{OH}^{-}\) produced

From the stoichiometry of the reaction, we know that 1 mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) will produce 2 moles of \(\mathrm{OH}^{-}\). Given the concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\), we can find out the amount of \(\mathrm{OH}^{-}\) produced: The concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\), 3.0 x \(10^{-7} \ \mathrm{M}\). To find the concentration of \(\mathrm{OH}^{-}\), we multiply the concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\) by 2, as per the balanced equation: \[ [\mathrm{OH}^{-}] = 2 \times [\mathrm{Ca}(\mathrm{OH})_{2}] \]
03

Calculate the concentration of \(\mathrm{OH}^{-}\)

By plugging in the given concentration of \(\mathrm{Ca}(\mathrm{OH})_{2}\), we can calculate the concentration of \(\mathrm{OH}^{-}\) ions: \[ [\mathrm{OH}^{-}] = 2 \times (3.0 \times 10^{-7} \ \mathrm{M})\] \[ [\mathrm{OH}^{-}] = 6.0 \times 10^{-7} \ \mathrm{M} \] The concentration of \(\mathrm{OH}^{-}\) ions in the solution is 6.0 x \(10^{-7}\ \mathrm{M}\).

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Most popular questions from this chapter

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

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