4.
a. Using Le Chatelier's principle, what effect will an increase of temperature have on the following system?
H2O(l) + 57 kJ ↔ H+(aq) + OH-(aq)
b. What is the expression for the constant for this reaction?
c. What effect will an increase in temperature have on the value of the constant? Why?
d. Justify the following statement: "A decrease in temperature decreases the value of Kwater"
5. What is the concentration of H+ in a solution containing 73.0 g of HCl in 0.5 L of water?
6. What is the [H+] of a solution obtained by adding 0.10 g of KOH to 1.0 L of water?
7. What are [H+] and [OH-] for 0.01 M perchloric acid?
8.
a. If [OH-] in a solution is 10-4 M, what is [H+]?
b. If 1500 mL of water is added to 500 mL of this solution, what will [H+] and [OH-] be?
4. a. An increase in temperature will shift the equilibrium to the left, favoring the reactants. This is because the forward reaction is exothermic and adding heat will cause the reaction to counteract the increase in temperature by producing more reactants.
b. The expression for the constant (equilibrium constant) can be written as: K = [H+(aq)][OH-(aq)] / [H2O(l)]. However, since water is in excess and its concentration remains approximately constant, it is usually omitted from the expression. Therefore, the expression for the constant in this reaction can be simplified to: K = [H+(aq)][OH-(aq)]
c. An increase in temperature will decrease the value of the equilibrium constant. This is because according to Le Chatelier's principle, when the temperature is increased, the equilibrium will shift in the direction that absorbs the excess heat, which in this case is the reverse reaction (forming H2O). This decrease in the concentration of H+(aq) and OH-(aq) will lead to a decrease in the value of the equilibrium constant, K.
d. The statement is incorrect. A decrease in temperature does not necessarily decrease the value of Kwater. The equilibrium constant for the reaction between water and its dissociation products (H+(aq) and OH-(aq)) is defined as Kwater = [H+(aq)][OH-(aq)]. Although temperature can affect the rate of the reaction or the concentrations of the species, it does not directly affect the value of the equilibrium constant Kwater.
5. To determine the concentration of H+ in the solution, we need to first calculate the number of moles of HCl present, and then divide it by the volume of the solution. The molar mass of HCl is 36.46 g/mol.
Number of moles of HCl = mass / molar mass = 73.0 g / 36.46 g/mol = 2.00 mol
Concentration of H+ = moles of H+ / volume of solution = 2.00 mol / 0.5 L = 4.00 M
Therefore, the concentration of H+ in the solution is 4.00 M.
6. To calculate the concentration of [H+], we need to first determine the number of moles of KOH added, and then divide it by the volume of the solution. The molar mass of KOH is 56.11 g/mol.
Number of moles of KOH = mass / molar mass = 0.10 g / 56.11 g/mol = 0.0018 mol
Concentration of H+ = moles of H+ / volume of solution = 0.0018 mol / 1.0 L = 0.0018 M
Therefore, the concentration of [H+] in the solution is 0.0018 M.
7. Perchloric acid (HClO4) is a strong acid that dissociates completely in water. Therefore, the concentration of [H+] will be equal to the initial concentration of HClO4, which is 0.01 M. The concentration of [OH-] can be calculated using the formula [H+][OH-] = 10^-14 M^2 (at 25°C).
[H+][OH-] = 10^-14 M^2
0.01 M * [OH-] = 10^-14 M^2
[OH-] = (10^-14 M^2) / 0.01 M
[OH-] = 10^-12 M
Therefore, [H+] is 0.01 M and [OH-] is 10^-12 M.
8. a. According to the definition of pH, it is the negative logarithm of the concentration of H+ ions.
pH = -log[H+]
Since [OH-] and [H+] are inversely related, we can use the formula: [H+][OH-] = 10^-14 M^2 (at 25°C).
[H+] = (10^-14 M^2) / [OH-]
[H+] = (10^-14 M^2) / 10^-4 M
[H+] = 10^-10 M
Therefore, [H+] is 10^-10 M.
8. b. When water is added to the solution, the volume increases, but the moles of the species present remain constant. Therefore, the concentration of [OH-] would remain the same, while the concentration of [H+] would decrease due to dilution.
[H+] after dilution = (initial moles of H+) / (total volume after dilution)
Initial moles of H+ = [H+] * initial volume
Initial moles of H+ = 10^-10 M * 500 mL = 5 * 10^-8 moles
Total volume after dilution = 500 mL + 1500 mL = 2000 mL = 2 L
[H+] after dilution = (5 * 10^-8 moles) / (2 L)
[H+] after dilution = 2.5 * 10^-8 M
[OH-] remains the same at 10^-4 M.
Therefore, [H+] after dilution is 2.5 * 10^-8 M and [OH-] remains 10^-4 M.
4a. To determine the effect of an increase in temperature on the given system, we can consider Le Chatelier's principle. According to this principle, if a stress (such as a change in temperature) is applied to a system at equilibrium, the system will shift in a way that minimizes the effect of the stress.
In this reaction, the forward reaction is endothermic because it absorbs energy in the form of heat (57 kJ). When the temperature is increased, it can be considered as adding energy to the system. To minimize this increase in energy, the equilibrium will shift in the direction that consumes heat.
Therefore, an increase in temperature will shift the equilibrium to the left, resulting in a decrease in the concentration of H+(aq) and OH-(aq), and an increase in the concentration of H2O(l).
4b. The expression for the equilibrium constant (K) for this reaction can be written as:
K = [H+(aq)][OH-(aq)] / [H2O(l)]
4c. An increase in temperature will cause a decrease in the value of the equilibrium constant (K). This is because the equilibrium constant is defined as the ratio of the concentrations of products to reactants at equilibrium. When the temperature increases, the forward reaction is favored because it consumes heat. This leads to a decrease in the concentration of products (H+(aq) and OH-(aq)) and an increase in the concentration of reactants (H2O(l)). As a result, the value of the equilibrium constant decreases.
4d. The given statement, "A decrease in temperature decreases the value of Kwater" is not accurate. The equilibrium constant for the reaction involving water, such as the dissociation of water into H+(aq) and OH-(aq), is always equal to 1 at a given temperature. Therefore, changing the temperature does not affect the value of Kwater.