Question

If the following solutions are mixed, is the resulting solution acidic, basic, or neutral? (a) \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HBr}\) and \(30.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH}\) (b) \(100.0 \mathrm{~mL}\) of \(0.0750 \mathrm{M} \mathrm{HCl}\) and \(75.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\).

Short answer

(a) The solution is basic. (b) The solution is basic.

Step-by-step solution

Calculate Moles of HBr

Convert the volume of HBr solution from milliliters to liters and then use the molarity to find the number of moles of HBr. \[ \text{L of HBr} = \frac{50.0}{1000} = 0.050 \text{ L} \]\[ \text{Moles of HBr} = 0.050 \times 0.100 = 0.0050 \text{ moles} \]

Calculate Moles of KOH

Similarly, convert the volume of KOH solution from milliliters to liters and use the molarity to find moles of KOH. \[ \text{L of KOH} = \frac{30.0}{1000} = 0.030 \text{ L} \]\[ \text{Moles of KOH} = 0.030 \times 0.200 = 0.0060 \text{ moles} \]

Determine Resulting Solution (Part a)

Compare the moles of HBr and KOH from Steps 1 and 2. Since there are fewer moles of HBr (0.0050) than KOH (0.0060), excess KOH remains after neutralization, making the solution basic.

Calculate Moles of HCl

Convert the volume of HCl solution from milliliters to liters and use the molarity to find the number of moles of HCl.\[ \text{L of HCl} = \frac{100.0}{1000} = 0.100 \text{ L} \] \[ \text{Moles of HCl} = 0.100 \times 0.0750 = 0.00750 \text{ moles} \]

Calculate Moles of Ba(OH)_2

Convert the volume of Ba(OH)\(_2\) solution from milliliters to liters and use the molarity to find the number of moles. Note that Ba(OH)\(_2\) provides two OH\(^-\) ions per molecule.\[ \text{L of Ba(OH)}_2 = \frac{75.0}{1000} = 0.075 \text{ L} \]\[ \text{Moles of Ba(OH)}_2 = 0.075 \times 0.100 = 0.00750 \text{ moles} \]\[ \text{OH}^- \text{ ions provided} = 2 \times 0.00750 = 0.0150 \text{ moles} \]

Determine Resulting Solution (Part b)

Compare the moles of HCl and OH\(^-\) ions from Step 4 and Step 5. Since there are 0.00750 moles of HCl and 0.0150 moles of OH\(^-\), the excess OH\(^-\) ions make the solution basic.

Key concepts

solution stoichiometry

Solution stoichiometry is a vital concept in chemistry, especially when dealing with reactions in solutions. It involves calculating the amounts of reactants and products in a chemical reaction using the concentration of solutions. This calculation ensures the correct stoichiometric proportions to predict the outcome of a chemical reaction.

In the example problems, solution stoichiometry is applied to determine the nature of the resulting solutions when mixing acids and bases. First, you need to know the molarity of each solution, which tells you how many moles of solute are present per liter of solution.

• For problem (a), calculate the moles of HBr and KOH to find the stoichiometric relation.
• Similarly, for problem (b), determine the moles of HCl and Ba(OH) 2 to establish the stoichiometry.

By using solution stoichiometry, you can identify the limiting reactant and determine whether the solution will be acidic, basic, or neutral.

molarity calculations

Molarity, often denoted as M, is a measurement of the concentration of a solution. It is defined by the number of moles of solute per liter of solution. Molarity allows us to precisely calculate the needed amounts of substances in various chemical reactions, particularly important in acid-base reactions.

To perform molarity calculations, first convert the volume from milliliters to liters, as molarity is expressed in terms of liters. Next, use the molarity equation: \[ ext{Moles of Solute} = ext{Volume (L)} imes ext{Molarity (M)}\]In the given exercises, this equation was used to find:

• Moles of HBr and KOH in part (a), and
• Moles of HCl and Ba(OH)2 in part (b).

Accurately calculating moles using molarity is crucial for predicting how much of each reactant will be consumed or remain after a reaction, helping to deduce whether the solution is acidic, basic, or neutral.

neutralization reactions

Neutralization reactions are specific types of chemical reactions where an acid and a base react to form water and a salt. These reactions are vital for determining the character of the resulting solution, whether it's acidic, basic, or neutral.

In the example exercises, neutralization involves the reaction between:

• HBr (acid) and KOH (base) in part (a), and
• HCl (acid) and Ba(OH) 2 (base) in part (b).

During these reactions, the hydrogen ions from the acid combine with the hydroxide ions from the base to form water. The products left over in excess after the reaction determine the nature of the solution:

• If hydroxide ions are in excess, the solution is basic.
• If hydrogen ions are in excess, the solution is acidic.

By calculating the moles and understanding stoichiometry, you can predict the resulting pH of a solution post-neutralization.