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Chapter 16: Problem 112

What is the pH of the solution obtained by mixing \(24.80 \mathrm{mL}\) of \(0.248 \mathrm{M} \mathrm{HNO}_{3}\) and \(15.40 \mathrm{mL}\) of \(0.394 \mathrm{M}\) \(\space\) \(KOH?\)

Short Answer

Expert verified
The pH of the solution obtained by mixing \(24.80 \mathrm{mL}\) of \(0.248 \mathrm{M} \mathrm{HNO}_{3}\) and \(15.40 \mathrm{mL}\) of $0.394 \mathrm{M}\( \)\space\( \)KOH$ is approximately 2.7. The solution is acidic due to the excess \( \mathrm{HNO}_3 \).

Step by step solution

01

Write the balanced chemical equation

The reaction between nitric acid (\( \mathrm{HNO}_3 \)) and potassium hydroxide (\( \mathrm{KOH} \)) can be written as follows: \( \mathrm{HNO}_3 + \mathrm{KOH} \rightarrow \mathrm{KNO}_3 + \mathrm{H}_2\mathrm{O} \). This equation tells us that one mole of acid reacts with one mole of base, so the stoichiometry of the reaction is 1:1.
02

Calculate the moles of each reactant

Next, we need to calculate the number of moles of \( \mathrm{HNO}_3 \) and \( \mathrm{KOH} \). The number of moles of a solution is calculated by multiplying the volume of the solution (in liters) by its molarity. For \( \mathrm{HNO}_3 \): Number of moles = \(0.0248 \mathrm{L} \times 0.248 \mathrm{M} = 0.00615 \mathrm{mol} \). For \( \mathrm{KOH} \): Number of moles = \(0.0154 \mathrm{L} \times 0.394 \mathrm{M} = 0.00607 \mathrm{mol} \).
03

Identify excess reactant

Because the stoichiometry of the reaction is 1:1, the reactant with the greater number of moles will be in excess. Here, \( \mathrm{HNO}_3 \) is in excess because it has more moles \( (0.00615 \) mol) than \( \mathrm{KOH} (0.00607\) mol). After reaction, there will be an excess of \( \mathrm{HNO}_3 \), and it is this excess \( \mathrm{HNO}_3 \) that determines the pH of the solution.
04

Calculate excess moles and final concentration

The excess amount of \( \mathrm{HNO}_3 \) is obtained by subtracting the moles of \( \mathrm{KOH} \) from moles of \( \mathrm{HNO}_3 \): Excess \( \mathrm{HNO}_3 \) = \(0.00615 \mathrm{mol} - 0.00607 \mathrm{mol} = 0.00008 \mathrm{mol} \). The total volume of the solution is \(24.80 \mathrm{mL} + 15.40 \mathrm{mL} = 40.20 \mathrm{mL} = 0.0402 \mathrm{L} \). The final concentration of \( \mathrm{HNO}_3 \) is then: \( \mathrm{concentration} = \frac{\mathrm{moles}}{\mathrm{volume}} = \frac{0.00008 \mathrm{mol}}{0.0402 \mathrm{L}} = 0.002 \mathrm{M} \).
05

Calculate pH

Now, calculate the pH of the solution. The formula for calculating pH is \( \mathrm{pH} = -\log[\mathrm{H^+}] \). For strong acids like \( \mathrm{HNO}_3 \), we can assume that they completely dissociate in water, so the concentration of \( \mathrm{H^+} \) ions is simply the concentration of the acid: \( \mathrm{pH} = -\log(0.002) = 2.7 \).
06

Conclusion

Therefore, the pH of the solution obtained by mixing \(24.80 \mathrm{mL}\) of \(0.248 \mathrm{M} \mathrm{HNO}_{3}\) and \(15.40 \mathrm{mL}\) of $0.394 \mathrm{M}\( \)\space\( \)KOH$ is approximately 2.7. Because the pH is less than 7, this tells us that the solution is acidic, which is expected because there was an excess of \( \mathrm{HNO}_3 \).

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Most popular questions from this chapter

Caproic acid, \(\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2},\) found in small amounts in coconut and palm oils, is used in making artificial flavors. A saturated aqueous solution of the acid contains \(11 \mathrm{g} / \mathrm{L}\) and has \(\mathrm{pH}=2.94 .\) Calculate \(K_{\mathrm{a}}\) for the acid. $$\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{O}_{2}^{-} \quad K_{\mathrm{a}}=?$$

\(\mathrm{CO}_{2}(\mathrm{g})\) 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, $$\mathrm{CO}_{2}(\mathrm{g})+\mathrm{LiOH}(\mathrm{s}) \longrightarrow \mathrm{LiHCO}_{3}(\mathrm{s})$$

The following four equilibria lie to the right: \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}+\) \(\mathrm{CH}_{3} \mathrm{NH}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}+\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ; \mathrm{H}_{2} \mathrm{SO}_{3}+\mathrm{F}^{-} \longrightarrow\) \(\mathrm{HSO}_{3}^{-}+\mathrm{HF} ; \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{NH}_{2}+\) \(\mathrm{H}_{2} \mathrm{O} ;\) and \(\mathrm{HF}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow \mathrm{F}^{-}+\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) (a) Rank all the acids involved in order of decreasing acid strength. (b) Rank all the bases involved in order of decreasing base strength. (c) State whether each of the following two equilibria lies primarily to the right or to the left: (i) \(\mathrm{HF}+\mathrm{OH}^{-} \longrightarrow \mathrm{F}^{-}+\mathrm{H}_{2} \mathrm{O} ;\) (ii) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\) \(\mathrm{HSO}_{3}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{SO}_{3}\).

Complete the following equations in those instances in which a reaction (hydrolysis) will occur. If no reaction occurs, so state. (a) \(\mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) (b) \(\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{NO}_{2}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) (c) \(\mathrm{K}^{+}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) (d) \(\mathrm{K}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq})+\) (e) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{O} \longrightarrow\)

How many milliliters of concentrated HCl(aq) \((36.0 \%\) HCl by mass, \(d=1.18 \mathrm{g} / \mathrm{mL}\) ) are required to produce \(12.5 \mathrm{L}\) of a solution with \(\mathrm{pH}=2.10 ?\)
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