Question

(a) What volume of \(0.115 \mathrm{M} \mathrm{HClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{M} \mathrm{NaOH} ?\) (b) What volume of \(0.128 \mathrm{M} \mathrm{HCl}\) is needed to neutralize \(2.87 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\) (c) If \(25.8 \mathrm{~mL}\) of \(\mathrm{AgNO}_{3}\) is needed to precipitate all the \(\mathrm{Cl}^{-}\) ions in a \(785-\mathrm{mg}\) sample of \(\mathrm{KCl}\) (forming \(\mathrm{AgCl}\), what is the molarity of the \(\mathrm{AgNO}_{3}\) solution? (d) If \(45.3 \mathrm{~mL}\) of \(0.108 \mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of \(\mathrm{KOH}\), how many grams of \(\mathrm{KOH}\) must be present in the solution?

Short answer

(a) 38.04 mL of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH. (b) 768.8 mL of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)2. (c) the molarity of the AgNO3 solution is 0.4081 M. (d) there are 0.2741 g of KOH in the solution.

Step-by-step solution

Write the balanced chemical equation

For the neutralization reaction between HClO4 and NaOH, the balanced equation is: \(HClO_{4}(aq) + NaOH(aq) -> NaClO_{4}(aq) + H_{2}O(l)\)

Calculate moles of NaOH

To find the moles of NaOH, we use the formula: \(Moles = Molarity \times Volume\) Moles of NaOH = 0.0875 M × 0.05000 L = 0.004375 mol

Calculate moles of HClO4

According to the balanced equation, one mole of HClO4 neutralizes one mole of NaOH. Thus, we need the same number of moles of HClO4 to neutralize the NaOH. Moles of HClO4 = 0.004375 mol

Calculate the volume of HClO4 solution

Now, we can find the volume of the HClO4 solution using the formula: \(Volume = \frac{Moles}{Molarity}\) Volume of HClO4 = \(\frac{0.004375 \ mol}{0.115 \ M}\) = 0.03804 L Convert the volume to milliliters: 0.03804 L × 1000 = 38.04 mL Answer: (a) 38.04 mL of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH. b) Find the volume of 0.128 M HCl needed to neutralize 2.87 g of Mg(OH)2

Write the balanced chemical equation

For the neutralization reaction between HCl and Mg(OH)2, the balanced equation is: \(Mg(OH)_{2}(s) + 2HCl(aq) -> MgCl_{2}(aq) + 2H_{2}O(l)\)

Convert grams of Mg(OH)2 to moles

First, find the molar mass of Mg(OH)2: \(Mg(OH)_{2} = 24.31 \ g/mol + 2 \times (16.00 \ g/mol + 1.01 \ g/mol) = 58.33 \ g/mol\) Now, convert 2.87 g of Mg(OH)2 to moles: \(\frac{2.87 \ g}{58.33 \ g/mol} = 0.04922 \ mol\)

Calculate moles of HCl needed

According to the balanced equation, 2 moles of HCl are needed to neutralize 1 mole of Mg(OH)2. Moles of HCl = 0.04922 mol of Mg(OH)2 × 2 = 0.09844 mol of HCl

Calculate the volume of HCl solution

Find the volume of the HCl solution using the formula: \(Volume = \frac{Moles}{Molarity}\) Volume of HCl = \(\frac{0.09844 \ mol}{0.128 \ M}\) = 0.7688 L Convert the volume to milliliters: 0.7688 L × 1000 = 768.8 mL Answer: (b) 768.8 mL of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)2. c) Calculate the molarity of AgNO3 solution needed to precipitate all Cl- ions in a 785 mg sample of KCl

Write the balanced chemical equation

For the precipitation reaction between KCl and AgNO3, the balanced equation is: \(KCl(aq) + AgNO_{3}(aq) -> AgCl(s) + KNO_{3}(aq)\)

Convert grams of KCl to moles

First, find the molar mass of KCl: \(KCl = 39.10 \ g/mol + 35.45 \ g/mol = 74.55 \ g/mol\) Now, convert 785 mg (0.785 g) of KCl to moles: \(\frac{0.785 \ g}{74.55 \ g/mol} = 0.010530 \ mol\)

Calculate moles of AgNO3 needed

According to the balanced equation, one mole of AgNO3 precipitates one mole of Cl- ions. Moles of AgNO3 = 0.010530 mol

Calculate the molarity of AgNO3 solution

Use the volume of the AgNO3 solution (25.8 mL = 0.0258 L) and the formula: \(Molarity = \frac{Moles}{Volume}\) Molarity of AgNO3 = \(\frac{0.010530 \ mol}{0.0258 \ L}\) = 0.4081 M Answer: (c) The molarity of the AgNO3 solution is 0.4081 M. d) Calculate the grams of KOH in a solution neutralized by 45.3 mL of 0.108 M HCl

Write the balanced chemical equation

For the neutralization reaction between KOH and HCl, the balanced equation is: \(KOH(aq) + HCl(aq) -> KCl(aq) + H_{2}O(l)\)

Calculate moles of HCl

Moles of HCl = Molarity × Volume = 0.108 M × 0.0453 L = 0.004885 mol

Calculate moles of KOH neutralized

According to the balanced equation, one mole of KOH is neutralized by one mole of HCl. Moles of KOH = 0.004885 mol

Calculate grams of KOH

First, find the molar mass of KOH: \(KOH = 39.10 \ g/mol + 15.999 \ g/mol + 1.008 \ g/mol = 56.108 \ g/mol\) Now, convert moles of KOH to grams: 0.004885 mol × 56.108 g/mol = 0.2741 g Answer: (d) There are 0.2741 g of KOH in the solution.

Key concepts

Stoichiometry

Stoichiometry is at the heart of chemical reactions. It's like a recipe that tells us how much of each substance is required to react completely with others. When working with stoichiometry, the first step is always to write down the balanced chemical equation. This shows the exact ratio of reactants and products involved in the reaction.

For instance, in a neutralization reaction where hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), the balanced stoichiometric equation would be:
\[ HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l) \]
Here, the equation tells us that one mole of HCl reacts with one mole of NaOH to produce one mole of sodium chloride (NaCl) and one mole of water (H_2O). Understanding this molar relationship is crucial for performing calculations related to the volumes and molarities of the reactants. Students sometimes assume that the coefficients in the balanced equation indicate the actual amounts of substances, but remember, they only represent the ratios. Always express the number of moles of substances present in those ratios to compute the results accurately.

Molarity and Volume Relationship

Molarity is a measure of concentration that expresses the number of moles of a solute per liter of solution. The relationship between molarity and volume is extremely important when solving problems related to solution chemistry. It's a straightforward concept: if we know the molarity of a solution and we have a certain volume of that solution, we can easily find out how many moles of solute we have.

The relationship is defined by the simple formula:
\[ Moles = Molarity \times Volume \]
This formula remains a constant guide through various calculations, whether you're working out how much acid is needed to neutralize a base or finding out the concentration of an unknown solution. This is effectively used in titration procedures where the concentration of one solution (the titrant) is used to determine the concentration of another solution (the analyte). Ensure to always convert the volume to liters when using this formula because molarity is defined as moles per liter. When solving problems, it's important not to forget units conversion – this can make or break the accuracy of your answers.

Precipitation Reactions

Precipitation reactions are a fascinating type of chemical reaction where two solutions react to form an insoluble solid called a precipitate. These reactions are visually compelling because you can often see the precipitate forming and settling out of the solution. It's like a magic show in the chemistry lab!

The key to understanding precipitation reactions is to remember that they occur when the ions of two reactants combine to form an insoluble compound. For example, when solutions of silver nitrate (AgNO3) and potassium chloride (KCl) are mixed, we get:
\[ AgNO_{3}(aq) + KCl(aq) \rightarrow AgCl(s) + KNO_{3}(aq) \]
Here, silver chloride (AgCl) is the precipitate. In textbook exercises, students might use this kind of reaction to find the molarity of one solution given the mass of the other substance using stoichiometry and the molarity-volume relationship discussed in previous sections. Remember though, it is essential to be aware of the solubility rules as they dictate which ionic compounds will precipitate under certain conditions. Problems involving precipitation reactions often ask students to calculate the amount of precipitate that will form or to determine the necessary concentrations to induce complete precipitation.