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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid-Base Reaction
An acid-base reaction involves the transfer of hydrogen ions from the acid to the base. In our example, nitric acid (\( \mathrm{HNO}_3 \)) is the acid and potassium hydroxide (\( \mathrm{KOH} \)) is the base. The balanced equation for the reaction is:
  • \( \mathrm{HNO}_3 + \mathrm{KOH} \rightarrow \mathrm{KNO}_3 + \mathrm{H}_2\mathrm{O} \)
This equation shows a 1:1 stoichiometry, meaning one mole of the acid reacts exactly with one mole of the base to form a salt (potassium nitrate) and water. In simple terms, when you mix equivalent amounts, they neutralize each other. If one is in excess, the solution remains either acidic or basic, depending on which is more abundant.
Stoichiometry
Stoichiometry helps us understand the quantitative relationships between reactants and products in a chemical reaction. In this scenario, the stoichiometry is 1:1, which means equal moles of \( \mathrm{HNO}_3 \) and \( \mathrm{KOH} \) will completely react. We calculate the moles of each reactant by multiplying their respective volumes (converted to liters) by their molarities:
  • \( \mathrm{HNO}_3 \) moles = \( 0.0248 \, \mathrm{L} \times 0.248 \, \mathrm{M} = 0.00615 \, \mathrm{mol} \)
  • \( \mathrm{KOH} \) moles = \( 0.0154 \, \mathrm{L} \times 0.394 \, \mathrm{M} = 0.00607 \, \mathrm{mol} \)
Since \( \mathrm{HNO}_3 \) has more moles, it is in excess, which is crucial for determining the final pH of the solution.
Molarity
Molarity (M) is a measure of the concentration of a solution expressed as the number of moles of solute per liter of solution. It gives us insights into how many moles of a substance are present in a given volume.To find the molarity of the excess \( \mathrm{HNO}_3 \) after the reaction, we calculate:
  • Excess moles of \( \mathrm{HNO}_3 \) = \( 0.00615 \, \mathrm{mol} - 0.00607 \, \mathrm{mol} = 0.00008 \, \mathrm{mol} \)
  • Total volume of solution = \( 24.80 \, \mathrm{mL} + 15.40 \, \mathrm{mL} = 40.20 \, \mathrm{mL} = 0.0402 \, \mathrm{L} \)
  • Molarity of excess \( \mathrm{HNO}_3 \) = \( \frac{0.00008 \, \mathrm{mol}}{0.0402 \, \mathrm{L}} = 0.002 \, \mathrm{M} \)
This molarity tells us how concentrated the excess acid is in the solution.
Chemical Equilibrium
Chemical equilibrium in the context of acid-base reactions involves the balance between the concentration of acids and bases. Here, because \( \mathrm{HNO}_3 \) is a strong acid, it dissociates completely, and the equilibrium heavily favors the products.When calculating the pH, we use the concentration of \( \mathrm{H^+} \) ions, which is equal to the concentration of \( \mathrm{HNO}_3 \) in this scenario. The pH is calculated by:
  • \( \mathrm{pH} = -\log[\mathrm{H^+}] \)
  • Substituting the available concentration: \( \mathrm{pH} = -\log(0.002) = 2.7 \)
A pH of 2.7 indicates an acidic solution, as expected from the excess nitric acid. Understanding equilibrium here underscores why the stronger acid dictates the pH.

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Most popular questions from this chapter

One handbook lists a value of 9.5 for \(\mathrm{p} \mathrm{K}_{\mathrm{b}}\) of quinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{N},\) a weak base used as a preservative for anatomical specimens and to make dyes. Another handbook lists the solubility of quinoline in water at \(25^{\circ} \mathrm{C}\) as \(0.6 \mathrm{g} / 100 \mathrm{mL} .\) Use this information to calculate the \(\mathrm{pH}\) of a saturated solution of quinoline in water.

Each of the following is a Lewis acid-base reaction. Which reactant is the acid, and which is the base? Explain. (a) \(\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\) (b) \(\operatorname{Zn}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})\)

Of the following, the amphiprotic ion is (a) \(\mathrm{HCO}_{3}^{-}\) (b) \(\mathrm{CO}_{3}^{2-} ;\) (c) \(\mathrm{NH}_{4}^{+} ;\) (d) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ;\) (e) \(\mathrm{ClO}_{4}^{-}\).

In \(0.10 \mathrm{M} \quad \mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{aq}), \quad\) (a) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.10 \mathrm{M}\) (b) \(\left[\mathrm{OH}^{-}\right]=0.10 \mathrm{M} ;(\mathrm{c}) \mathrm{pH}<7 ;(\mathrm{d}) \mathrm{pH}<13\).

What mass of benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), would you dissolve in \(350.0 \mathrm{mL}\) of water to produce a solution with a \(\mathrm{pH}=2.85 ?\) $$\begin{aligned} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} & \\ K_{\mathrm{a}}=6.3 \times 10^{-5} \end{aligned}$$
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