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Chapter 16: Problem 19

\(50.00 \mathrm{mL}\) of \(0.0155 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) is mixed with \(75.00 \mathrm{mL}\) of 0.0106 M KOH(aq). What is the pH of the final solution?

Short Answer

Expert verified
The pH of the final solution is 10.2

Step by step solution

01

Calculate the number of moles for HI and KOH

Using the formula Molarity (M)= moles of solute/liters of solution, calculate moles of HI and KOH. Moles of HI = Molarity * Volume = 0.0155 M * 0.050 L = \(7.75 \times 10^{-4} \, mol\). Moles of KOH = 0.0106 M * 0.075 L = \(7.95 \times 10^{-4} \, mol\)
02

Determine the limiting reactant

In this case, HI is the limiting reactant and KOH is the excess reactant. This is because the moles of HI is less than the moles of KOH.
03

Determine the number of remaining moles of KOH

Calculate the moles of KOH left after reaction with HI. This is obtained by subtracting the moles of HI from the moles of KOH: \(7.95 \times 10^{-4} mol \, - \, 7.75 \times 10^{-4} mol = 2 \times 10^{-5} mol\, KOH\)
04

Calculate the OH- ion concentration in the solution

Since KOH is a strong base, it dissociates completely into K+ and OH- in solution. Therefore, the concentration of OH- ions is equal to the concentration of KOH. The volume of the solution is the sum of the volumes of HI and KOH, which is 0.050 L + 0.075 L = 0.125 L. The concentration \(OH^- = \frac { moles of OH^- }{Volume of Solution } = \frac {2 \times 10^{-5} mol} { 0.125 L} = 1.6 \times 10^{-4} M).
05

Calculate the pOH and then the pH of the solution

The pOH is calculated as \( -\log[OH^-] \, = \, -\log(1.6 \times 10^{-4}) \, = \, 3.8 \). The pH is then calculated as \(14 - pOH \, = \, 14 - 3.8 = 10.2\)

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Most popular questions from this chapter

A 625 mL sample of an aqueous solution containing 0.275 mol propionic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H},\) has \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.00239 \mathrm{M} .\) What is the value of \(K_{\mathrm{a}}\) for propionic acid? $$\begin{aligned} &\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2}\\\ &&K_{\mathrm{a}}=? \end{aligned}$$

A 28.2 L volume of \(\mathrm{HCl}(\mathrm{g}),\) measured at \(742 \mathrm{mmHg}\) and \(25.0^{\circ} \mathrm{C},\) is dissolved in water. What volume of \(\mathrm{NH}_{3}(\mathrm{g}),\) measured at \(762 \mathrm{mmHg}\) and \(21.0^{\circ} \mathrm{C},\) must be absorbed by the same solution to neutralize the HCl?

Explain the important distinctions between each pair of terms: (a) Bronsted- Lowry acid and base; (b) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH} ;\) (c) \(K_{\mathrm{a}}\) for \(\mathrm{NH}_{4}^{+}\) and \(K_{\mathrm{b}}\) for \(\mathrm{NH}_{3} ;\) (d) leveling effect and electron- withdrawing effect.

\(\mathrm{CO}_{2}(\mathrm{g})\) can be removed from confined quarters (such as a spacecraft) by allowing it to react with an alkali metal hydroxide. Show that this is a Lewis acid-base reaction. For example, $$\mathrm{CO}_{2}(\mathrm{g})+\mathrm{LiOH}(\mathrm{s}) \longrightarrow \mathrm{LiHCO}_{3}(\mathrm{s})$$

For each reaction draw a Lewis structure for each species and indicate which is the acid and which is the base: (a) \(\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{O}+\mathrm{BF}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{OBF}_{3}\) (c) \(\mathrm{O}^{2-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{OH}^{-}\) (d) \(\mathrm{S}^{2-}+\mathrm{SO}_{3} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}\)
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