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Chapter 4: Problem 21

What is the molarity of a \(\mathrm{HNO}_{3}\) solution if \(68.5 \mathrm{~mL}\) is needed to react with \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M}\) KOH solution? The equation is $$ \mathrm{HNO}_{3}(a q)+\mathrm{KOH}(a q) \longrightarrow \mathrm{KNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$

Short Answer

Expert verified
The molarity of the \( \mathrm{HNO}_3 \) solution is \( 0.0547 \mathrm{M} \).

Step by step solution

01

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction given is \( \mathrm{HNO}_{3}(aq) + \mathrm{KOH}(aq) \rightarrow \mathrm{KNO}_{3}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \). This shows us that 1 mole of \( \mathrm{HNO}_3 \) reacts with 1 mole of \( \mathrm{KOH} \) to produce 1 mole of \( \mathrm{KNO}_3 \) and water.
02

Calculate Moles of KOH

Use the molarity and volume of the KOH solution to find the number of moles of KOH. The formula is: \( \text{Moles of KOH} = \text{Molarity} \times \text{Volume (in liters)} \). Substituting the values: \( 0.150 \mathrm{~M} \times 0.0250 \mathrm{~L} = 0.00375 \) moles.
03

Relate Moles of HNO₃ to Moles of KOH

From the balanced equation, 1 mole of \( \mathrm{HNO}_3 \) reacts with 1 mole of \( \mathrm{KOH} \). Thus, the moles of \( \mathrm{HNO}_3 \) required is equal to the moles of \( \mathrm{KOH} \), which is \( 0.00375 \) moles.
04

Calculate Molarity of HNO₃ Solution

The molarity equation is \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). Substitute the respective values to find molarity: \( \frac{0.00375 \text{ moles}}{0.0685 \text{ L}} = 0.0547 \mathrm{M} \).

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Key Concepts

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

Acid-Base Titration
Acid-base titration is a method used in chemistry to determine the concentration of an acid or base in a solution. This technique involves slowly adding a titrant, which is a solution of known concentration, to a solution containing the analyte, whose concentration is unknown, until the reaction reaches its equivalence point. The equivalence point is the stage where the acid completely neutralizes the base (or vice versa), forming water and a salt.
In the given exercise, nitric acid (\( \mathrm{HNO}_{3} \)) is titrated with potassium hydroxide (\( \mathrm{KOH} \)). By measuring the volume of \( \mathrm{HNO}_{3} \) needed to reach neutralization with the known molarity and volume of \( \mathrm{KOH} \), we can calculate the molarity of the \( \mathrm{HNO}_{3} \) solution. This practical application of acid-base titration is often used in labs to analyze the concentration of acidic or basic solutions.
Balanced Chemical Equation
Balanced chemical equations are essential for understanding chemical reactions. A balanced equation has the same number of each type of atom on both sides of the reaction, maintaining the principle of the conservation of mass.
In the exercise, the balanced chemical equation is given as: \( \mathrm{HNO}_{3}(aq) + \mathrm{KOH}(aq) \rightarrow \mathrm{KNO}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \). This equation indicates that one molecule of nitric acid reacts with one molecule of potassium hydroxide to produce one molecule of potassium nitrate and one molecule of water.
Understanding balanced equations helps us predict how much of each reactant is needed and how much product will be formed. In this case, knowing that the mole ratio is 1:1 between \( \mathrm{HNO}_{3} \) and \( \mathrm{KOH} \) simplifies the stoichiometry calculations.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It helps chemists determine the amounts of substances consumed and produced in a reaction.
In the problem, stoichiometry is used to link the volume and molarity of the \( \mathrm{KOH} \) solution to the nitric acid. First, we calculated the moles of \( \mathrm{KOH} \) using its molarity and volume: \( 0.150 \mathrm{~M} \times 0.0250 \mathrm{~L} = 0.00375 \text{ moles of } \mathrm{KOH} \).
Then, from the balanced equation, we know that this number of moles is equal to the moles of \( \mathrm{HNO}_{3} \) needed. Finally, the molarity of \( \mathrm{HNO}_{3} \) was calculated using the formula:
  • \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)
  • \( M = \frac{0.00375 \text{ moles}}{0.0685 \text{ L}} = 0.0547 \mathrm{M} \).
Understanding stoichiometry is crucial for calculating reactant or product quantities accurately in any chemical reaction.

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Most popular questions from this chapter

Which element is oxidized and which is reduced in each of the following reactions? (a) \(\mathrm{Ca}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{Sn}(s)\) (b) \(\mathrm{ICl}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCl}(a q)+\mathrm{HOI}(a q)\)

A flask containing \(450 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) HBr was accidentally knocked to the floor. How many grams of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) would you need to put on the spill to neutralize the acid according to the following equation? \(2 \mathrm{HBr}(a q)+\mathrm{K}_{2} \mathrm{CO}_{3}(a q) \longrightarrow 2 \mathrm{KBr}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\).

The solubility of an ionic compound can be described quantitatively by a value called the solubility product constant, \(K_{s \mathrm{p}}\). For the general solubility process \(\mathrm{A}_{a} \mathrm{~B}_{b} \rightleftharpoons a \mathrm{~A}^{n+}+b \mathrm{~B}^{m-}, K_{\mathrm{sp}}=\left[\mathrm{A}^{n+}\right]^{a}\left[\mathrm{~B}^{\mathrm{m}-}\right]^{b} .\) The brackets refer to concentrations in moles per liter. (a) Write the expression for the solubility product constant of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) (b) If \(K_{\text {sp }}=1.1 \times 10^{-12}\) for \(\mathrm{Ag}_{2} \mathrm{Cr} \mathrm{O}_{4}\), what are the molar concentrations of \(\mathrm{Ag}^{+}\) and \(\mathrm{CrO}_{4}^{2-}\) in solution?

Assume that you have an aqueous mixture of \(\mathrm{BaCl}_{2}\) and \(\mathrm{CuCl}_{2}\). How could you use a precipitation reaction to separate the two metal ions?

(a) Use the following reactions to arrange the elements \(\mathbf{A}, \mathbf{B}, \mathbf{C}\), and \(\mathbf{D}\) in order of their decreasing ability as reducing agents: \(2 \mathrm{~A}+\mathrm{B}^{2+} \longrightarrow 2 \mathrm{~A}^{+}+\mathrm{B} \quad \mathrm{B}+\mathrm{D}^{2+} \longrightarrow \mathrm{B}^{2+}+\mathrm{D}\) \(\mathrm{A}^{+}+\mathrm{C} \longrightarrow\) no reaction \(2 \mathrm{C}+\mathrm{B}^{2+} \longrightarrow 2 \mathrm{C}^{+}+\mathrm{B}\) (b) Which of the following reactions would you expect to occur according to the activity series you established in part (a)? (1) \(2 \mathrm{~A}^{+}+\mathrm{D} \longrightarrow 2 \mathrm{~A}+\mathrm{D}^{2+}\) (2) \(\mathrm{D}^{2+}+2 \mathrm{C} \longrightarrow \mathrm{D}+2 \mathrm{C}^{+}\)
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