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Chapter 14: Problem 88

Calculate \(\left[\mathrm{OH}^{-}\right], \mathrm{pOH}\), and \(\mathrm{pH}\) for each of the following. a. \(0.00040 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) b. a solution containing \(25 \mathrm{~g}\) KOH per liter c. a solution containing \(150.0 \mathrm{~g} \mathrm{NaOH}\) per liter

Short Answer

Expert verified
a. \(\left[\mathrm{OH}^{-}\right] = 0.00080 \mathrm{M}\), \(\mathrm{pOH} = 3.10\), \(\mathrm{pH} = 10.90\) b. \(\left[\mathrm{OH}^{-}\right] = 0.64 \mathrm{M}\), \(\mathrm{pOH} = 0.19\), \(\mathrm{pH} = 13.81\) c. \(\left[\mathrm{OH}^{-}\right] = 3.75 \mathrm{M}\), \(\mathrm{pOH} = -0.57\), \(\mathrm{pH} = 14.57\)

Step by step solution

01

Determining the concentration of \(\mathrm{OH}^{-}\) ions in solution

Since \(1 \mathrm{Ca}(\mathrm{OH})_{2}\) molecule dissociates into \(2 \mathrm{OH}^-\) ions in aqueous solutions, we have: \(\left[\mathrm{OH}^{-}\right] = 2 \times 0.00040 \mathrm{M} = 0.00080 \mathrm{M}\).
02

Calculate pOH

To find the pOH, we use the formula \(\mathrm{pOH}=-\log_{10}\left[\mathrm{OH}^{-}\right]\). So, \(\mathrm{pOH} = -\log_{10}(0.00080) = 3.10\).
03

Calculate pH

We can now find the pH using the formula \(\mathrm{pH} = 14 - \mathrm{pOH}\). Therefore, \(\mathrm{pH} = 14 - 3.10 = 10.90\). b. a solution containing \(25 \mathrm{~g}\) KOH per liter
04

Determine the concentration of \(\mathrm{OH}^{-}\) ions in solution

First, we need to find the molar concentration of KOH in the solution. The molar mass of KOH is approximately \(39.10 \mathrm{~g/mol}\). So, the molar concentration is \(\frac{25 \mathrm{~g}}{39.10 \mathrm{~g/mol}} = 0.64 \mathrm{M}\). Since every KOH molecule forms one \(\mathrm{OH}^-\) ion, the concentration of \(\left[\mathrm{OH}^{-}\right] = 0.64 \mathrm{M}\).
05

Calculate pOH

Using the formula \(\mathrm{pOH}=-\log_{10}\left[\mathrm{OH}^{-}\right]\), the pOH is \(\mathrm{pOH} = -\log_{10}(0.64) = 0.19\).
06

Calculate pH

To find the pH, use the formula \(\mathrm{pH} = 14 - \mathrm{pOH}\). Thus, \(\mathrm{pH} = 14 - 0.19 = 13.81\). c. a solution containing \(150.0 \mathrm{~g} \mathrm{NaOH}\) per liter
07

Determine the concentration of \(\mathrm{OH}^{-}\) ions in solution

First, find the molar concentration of NaOH in the solution. The molar mass of NaOH is about \(40.00 \mathrm{~g/mol}\). So, the molar concentration is \(\frac{150.0 \mathrm{~g}}{40.00 \mathrm{~g/mol}} = 3.75 \mathrm{M}\). Since every NaOH molecule forms one \(\mathrm{OH}^-\) ion, the concentration of \(\left[\mathrm{OH}^{-}\right] = 3.75 \mathrm{M}\).
08

Calculate pOH

Using the formula \(\mathrm{pOH}=-\log_{10}\left[\mathrm{OH}^{-}\right]\), the pOH is \(\mathrm{pOH} = -\log_{10}(3.75) = -0.57\).
09

Calculate pH

To find the pH, use the formula \(\mathrm{pH} = 14 - \mathrm{pOH}\). Thus, \(\mathrm{pH} = 14 - (-0.57) = 14.57\).

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Most popular questions from this chapter

A sample containing \(0.0500 \mathrm{~mol} \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is dissolved in enough water to make \(1.00 \mathrm{~L}\) of solution. This solution contains hydrated \(\mathrm{SO}_{4}{ }^{2-}\) and \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) ions. The latter behaves as an acid: $$\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}^{+}(a q)$$ a. Calculate the expected osmotic pressure of this solution at \(25^{\circ} \mathrm{C}\) if the above dissociation is negligible. b. The actual osmotic pressure of the solution is \(6.73\) atm at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{a}}\) for the dissociation reaction of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\). (To do this calculation, you must assume that none of the ions goes through the semipermeable membrane. Actually, this is not a great assumption for the tiny \(\mathrm{H}^{+}\) ion.)

Calculate the \(\mathrm{pH}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The prin- cipal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\)

Use the Lewis acid-base model to explain the following reaction. $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$

Monochloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}\), is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of \(K_{\mathrm{a}}\) for monochloroacetic acid is \(1.35 \times 10^{-3}\). Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of monochloroacetic acid.

Write out the stepwise \(K_{\mathrm{a}}\) reactions for the diprotic acid \(\mathrm{H}_{2} \mathrm{SO}_{3}\).
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