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Chapter 14: Problem 93

What mass of KOH is necessary to prepare 800.0 \(\mathrm{mL}\) of a solution having a \(\mathrm{pH}=11.56 ?\)

Short Answer

Expert verified
The mass of KOH needed to prepare 800.0 mL of a solution having a pH of 11.56 is \(0.163 g\).

Step by step solution

01

Determine the concentration of the KOH solution using the given pH value.

The pH value determines the concentration of hydrogen ions (H+) in the solution. Since KOH is a strong base, the pH value provided gives us information about the pOH value, which determines the concentration of hydroxide ions (OH-). The relationship between pH and pOH is: \[pH + pOH = 14\] First, we need to find the pOH value: \: pOH = 14 - pH Then, we can use the pOH value to find the concentration of the OH- ions in the solution: \[ [OH^-] = 10^{-pOH} \]
02

Calculate the concentration of the KOH solution.

Now, we can calculate the pOH value and use it to determine the concentration of the OH- ions in the solution. \ \: pOH = 14 - 11.56 = 2.44 \ \: [OH^-] = 10^{-2.44} = 3.63 × 10^{-3} M Since KOH is a strong base and it dissociates completely in water, the concentration of the KOH solution is equal to the concentration of OH-. Therefore, the concentration of the KOH solution is 3.63 ×10^{-3} M.
03

Use the volume of the solution to find the moles of KOH needed.

To find the number of moles of KOH required for the given volume, we can use the formula: \[moles\ of\ KOH = Volume\ of\ solution × Concentration\ of\ solution\] We have the volume of the solution in mL, so we need to convert it to liters: \[ Volume\ of\ solution = 800.0 mL × \frac{1\: L}{1000\: mL} = 0.800 L \] Now, we can calculate the moles of KOH: \: moles\ of\ KOH = 0.800 L × 3.63 × 10^{-3} M = 2.904 × 10^{-3} mol
04

Calculate the mass of KOH required.

Lastly, we will use the molar mass of KOH to convert moles to mass. The molar mass of KOH is: \[Molar\ mass\ of\ KOH\ = 39.1\ (K) + 15.999\ (O) + 1.008\ (H) = 56.1\: g/mol\] Now, we can calculate the mass of KOH required: \[mass\ of\ KOH = moles\ of\ KOH × Molar\ mass\ of\ KOH\] \: mass\ of\ KOH = 2.904 × 10^{-3} mol × 56.1 g/mol = 0.163 g
05

Final Answer:

The mass of KOH needed to prepare 800.0 mL of a solution having a pH of 11.56 is 0.163 g.

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Most popular questions from this chapter

Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. \(\mathrm{Li}_{2} \mathrm{O}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{SrO}\)

Place the species in each of the following groups in order of increasing base strength. Give your reasoning in each case. a. \(\mathrm{IO}_{3}^{-}, \mathrm{BrO}_{3}^{-}\) b. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}\) c. \(\mathrm{OCl}^{-}, \mathrm{OI}^{-}\)

The \(\mathrm{pH}\) of a \(0.016-M\) aqueous solution of \(p\) -toluidine \(\left(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2}\right)\) is 8.60 . Calculate \(K_{\mathrm{b}}\) .

An aqueous solution contains a mixture of 0.0500\(M \mathrm{HCOOH}\) \(\left(K_{\mathrm{a}}=1.77 \times 10^{-4}\right)\) and 0.150$M \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\left(K_{\mathrm{a}}=\right.\( \)1.34 \times 10^{-5} ) .\( Calculate the \)\mathrm{pH}$ of this solution. Because both acids are of comparable strength, the \(\mathrm{H}^{+}\) contribution from both acids must be considered.

Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced chemical equations for the reactions causing the solution to be acidic or basic. The relevant \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values are found in Tables 14.2 and \(14.3 .\) $\begin{array}{ll}{\text { a. NaNO, }} & {\text { d. } \mathrm{NH}_{4} \mathrm{NO}_{2}} \\ {\text { b. NaNO_ }_{2}} & {\text { e. } \mathrm{KOCl}} \\\ {\text { c. } \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHClO}_{4}} & {\text { f. } \mathrm{NH}_{4} \mathrm{OCl}}\end{array}$
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