Chemistry Answer Note #13

Question 1 Suppose that A and B react to form C according to the equation below. A + 2 B ⇌ C What are the equilibrium concentrations of A, B and C if 2.8 mol A and 2.8 mol B are added to a 1.0 L flask? Assume that the equilibrium constant for this reaction is Kc = 3.3x1011. Hint: The equilibrium constant for this reaction is quite large. Therefore, the reaction goes almost 100% to completion. Enter your answers with two significant figures. Do not include units as part of your answer. Use exponential notation (e.g. enter 1.2x10-10 as 1.2e-10; enter 1.2x1015 as 1.2e15.) Equilibrium concentration of A: Number mol L−1 Equilibrium concentration of B: Number mol L−1 Equilibrium concentration of C: Number mol L−1 Answer The equilibrium concentrations of A, B, and C can be determined using the equilibrium constant expression, Kc = [C]/([A][B]^2). First, let us calculate the initial concentrations of A and B in the flask. Since 2.8 mol of each reactant is added to a 1.0 L flask, the initial concentration of A and B will be 2.8 mol/L. Now, since the equilibrium constant Kc is quite large (3.3x1011), this indicates that the reaction goes almost 100% to completion. This means that the product, C, will have a significantly higher concentration compared to the reactants, A and B. Using the equilibrium constant expression, we can rearrange it to solve for the equilibrium concentration of C, [C] = Kc x [A] x [B]^2. Plugging in the values, we get [C] = (3.3x1011) x (2.8 mol/L) x (2.8 mol/L)^2 = 7.04x1012 mol/L. Therefore, the equilibrium concentration of C is 7.04x1012 mol/L. Since the reaction goes almost to completion, the equilibrium concentrations of A and B will be significantly lower. To calculate these, we can use the equilibrium constant expression again and solve for [A] and [B]. This gives us [A] = [B] = 5.33x10^-12 mol/L. Question 2 8.61 mL of 0.245 M HBr was needed to neutralize a 5.00 mL sample of Ca(OH)2. What was the initial Ca(OH)2 concentration? Answer To answer this question, we need to use the equation M1V1 = M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume. Firstly, we need to convert the given volume and concentration of HBr into molarity. This can be done by dividing the given concentration (0.245 M) by the molar mass of HBr (80.91 g/mol). This gives us a final molarity of 3.03 x 10^-3 M. Next, we can use the given values of V1 and M2 to find the initial concentration of Ca(OH)2. Plugging in the values into the equation, we get: (0.245 M)(8.61 mL) = M2(5.00 mL) M2 = (0.245 M)(8.61 mL) / 5.00 mL M2 = 0.422 M Therefore, the initial concentration of Ca(OH)2 is 0.422 M. Question 3 Given the following saponification reaction of an ester with sodium hydroxide: CH3COOCH3+OH- yields CH3COO2^- + CH3OH. The reaction is carried out at initial concentrations of 1.10^-2 of reactants and at a temperature such that 1% of the ester is saponified in 30seconds. How long will 50% of the ester be saponified? Answer The given saponification reaction involves the hydrolysis of an ester, CH3COOCH3, with sodium hydroxide, leading to the formation of a carboxylate ion, CH3COO2^- , and an alcohol molecule, CH3OH. This reaction is commonly referred to as the alkaline hydrolysis of esters and is widely used in the production of soaps and detergents. In this answer, we will be discussing the factors affecting the rate of this reaction and applying them to determine the time required for 50% of the ester to be saponified. Firstly, let's consider the initial concentrations of the reactants, which are 1.10^-2. This means that at the start of the reaction, there are 1.10^-2 moles of both the ester and the sodium hydroxide for every liter of the reaction mixture. The higher the concentration of the reactants, the faster the reaction rate will be. This is because with a higher concentration, there are more particles available to collide and react, resulting in a higher frequency of successful collisions. Next, we must take into account the temperature at which the reaction is carried out. As mentioned in the question, the reaction is performed at a temperature such that 1% of the ester is saponified in 30 seconds. This temperature can be determined using the Arrhenius equation, which relates the rate constant of a reaction to the activation energy and temperature. However, since we are not provided with the values for the activation energy and rate constant, we cannot accurately determine the temperature using this method. Alternatively, we can use the Arrhenius plot method, which involves plotting the natural logarithm of the reaction rate against the reciprocal of the temperature. The point where this plot intersects the y-axis corresponds to the activation energy of the reaction while the slope of the line represents the rate constant. This method is more time-consuming but can yield more accurate results. Now, to determine the time required for 50% of the ester to be saponified, we can use the first-order integrated rate law for the reaction. This law states that the rate of a first-order reaction is directly proportional to the concentration of the reactant over time. Using this, we can set up the following equation: ln([ester]t/[ester]0) = -kt Where [ester]t is the concentration of the ester at time t, [ester]0 is the initial concentration of the ester, k is the rate constant, and t is the time taken for the reaction to occur. Rearranging this equation, we get: t = ln([ester]t/[ester]0)/(-k) Since the initial concentration of the ester and sodium hydroxide are the same, we can substitute the given values (1.10^-2) for [ester]t/[ester]0. To determine the value of k, we can use the fact that 1% of the ester is saponified in 30 seconds, which means that at t = 30 seconds, [ester]t = 0.01[ester]0. Substituting all the values, we get: t = ln(0.01)/(-k) = 72.75 seconds Therefore, it would take approximately 72.75 seconds for 50% of the ester to be saponified under these conditions. In conclusion, the time required for 50% of the ester to be saponified depends on the initial concentrations of the reactants and the temperature at which the reaction is carried out. By using the first-order integrated rate law and the given information, we were able to determine the time required for this reaction. For further understanding, I would recommend referring to the following sources: 1. "Organic Chemistry" by David R. Klein 2. "Physical Chemistry" by Peter Atkins and Julio de Paula 3. "Chemical Kinetics and Reaction Dynamics" by Santosh K. Upadhyay 4. "Introduction to Chemical Kinetics" by Margaret Robson Wright 5. "Saponification" by Encyclopaedia Britannica These sources provide detailed explanations and examples of reaction kinetics and saponification reactions. Additionally, they also cover topics such as the Arrhenius equation and the Arrhenius plot method, which were mentioned in this answer. With a deeper understanding of the concepts discussed in these sources, one can solve complex problems and calculations in regards to chemical reactions with greater ease and accuracy. Question 4 A mono basic acid has Ka = 1. 8 × 10−5 at 25 C. Calculate its degree of dissociation at a concentration of 0.2M at the same temperature. What will be the concentration of [H] + furnished by the acid ? Also calculate the pH of the solution. Answer A mono basic acid is an acid that can donate only one proton (H+) per molecule. Ka, also known as the acid dissociation constant, is a measure of the strength of an acid. It is defined as the ratio of products to reactants at equilibrium for a weak acid dissociation reaction. In this case, we are given that Ka = 1.8 × 10−5 at a temperature of 25 C and we need to find the degree of dissociation at a concentration of 0.2M. To calculate the degree of dissociation, we can use the expression: α = √(Ka/[H+]) Where: α = degree of dissociation Ka = acid dissociation constant [H+] = concentration of hydrogen ions (protons) Substituting the given values, we get: α = √(1.8 × 10−5 / 0.2M) α = √(9 × 10^-5) α = 0.00949 or 0.95% This means that the degree of dissociation of the acid is approximately 0.95%. Moving on to the next part of the question, we need to find the concentration of [H+] furnished by the acid. The concentration of [H+] can be calculated using the following expression: [H+] = √(Ka x [HA]) Where: HA = initial concentration of the acid Substituting the given values, we get: [H+] = √(1.8 × 10−5 × 0.2M) [H+] = √3.6 × 10−6 [H+] = 6 × 10−3M or 0.006M Therefore, the concentration of [H+] furnished by the acid at a concentration of 0.2M is 0.006M or 6 × 10−3M. Finally, we need to calculate the pH of the solution. pH is a measure of the acidity or basicity of a solution and is defined as the negative logarithm of the hydrogen ion concentration. pH = -log[H+] Substituting the value for [H+], we get: pH = -log(0.006) pH = 2.22 Therefore, the pH of the solution is 2.22. Question 5 Is an aqueous solution of CO2 acidic, basic or neutral? Include an equation to illustrate this behaviour. (e.g., if acidic, the equation will have H+(aq) as a product and if basic, the equation will have OH–(aq) as a product.) Answer According to the Arrhenius theory, an acidic solution contains an excess of hydrogen ions (H+) and a basic solution contains an excess of hydroxide ions (OH-). A neutral solution has an equal concentration of H+ ion and OH- ion, resulting in a pH of 7. With this in mind, let's dive into the behaviour of an aqueous solution of CO2. CO2 is a non-metal oxide which dissolves in water to form carbonic acid (H2CO3). This acid then dissociates to release H+ ions, giving the solution an acidic nature. The equation for the dissociation of carbonic acid is given by: H2CO3(aq) ⇌ H+(aq) + HCO3-(aq) The release of H+ ions in this reaction confirms that the solution is indeed acidic. In fact, carbonic acid is a weak acid, meaning that it does not completely dissociate in water and thus, does not release a large number of H+ ions. This is why an aqueous solution of CO2 is considered to be a weak acid. Now, let's look at the pH of the solution. pH is a scale that measures the acidity or basicity of a solution, with values ranging from 0 (most acidic) to 14 (most basic). A neutral solution has a pH of 7, and anything below 7 is considered acidic. Using Le Chatelier's principle, we can deduce that the addition of CO2 to water will result in the formation of more H+ ions, lowering the pH of the solution and making it more acidic. To determine the pH of an aqueous solution of CO2, we can use the HendersonHasselbalch equation, which is given by: pH = pKa + log([HCO3-]/[CO2]) where pKa is the acid dissociation constant for carbonic acid and [HCO3-] and [CO2] are the concentrations of bicarbonate and carbon dioxide, respectively. If we assume that the pKa of carbonic acid is 6.35 and the concentration of HCO3- is equal to the concentration of CO2 (since the reaction has not proceeded far), we can rearrange the Henderson-Hasselbalch equation to get: pH = 6.35 + log(1/1) pH = 6.35 Therefore, an aqueous solution of CO2 will have a pH of approximately 6.35, making it slightly acidic.