CH3NH2(aq)+H2O(l)=>CH3NH3+(aq)+OH-(aq) Kb=4.4 x 10^-4
Methylamine, CH3NH2, is a weak base that reacts with water according to the equation above. A student obtains a 50.0 mL sample of a methylamine solution and determines the pH of the solution to be 11.77.
(a) Write the expression for the equilibrium constant, Kb, for methylamine.
(b) Calculate the molar concentration of OH- in the 50.0 mL sample of the methylamine solution.
(c) Calculate the initial molar concentration of CH3NH2(aq) in the solution before it reacted with water and equilibrium was established.
The 50.0 mL sample of the methylamine solution is titrated with HCl solution of unknown concentration. The equivalence point of the titration is reached after a volume of 36.0 mL of the HCl solution is added. The pH of the solution at the equivalence point is 5.98.
(d) Write the net-ionic equation that represents the reaction that takes place during the titration.
(e) Calculate the concentration of the HCl solution used to titrate the methylamine.
(f) using the axes provided, sketch the titration curve that results from the titration described above. On the graph, clearly label the equivalence point of the titration.
(a) Kb = [CH3NH3+][OH-] / [CH3NH2]
(b) Kb = 4.4 x 10^-4
[OH-] = Kb x [CH3NH2] = 4.4 x 10^-4 x 0.050 L = 2.2 x 10^-5 M
(c) [CH3NH2] = [OH-] / Kb = 2.2 x 10^-5 / 4.4 x 10^-4 = 0.05 M
(d) CH3NH2(aq) + HCl(aq) → CH3NH3+(aq) + Cl-(aq)
(e) [HCl] = [OH-] / (Vtitrate/Vsample) = 2.2 x 10^-5 / (36.0/50.0) = 1.5 x 10^-4 M
(f) See graph below. The equivalence point is at pH 5.98.
(a) The expression for the equilibrium constant, Kb, for methylamine is:
\[K_b = \frac{[\text{CH}_3\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{NH}_2]}\]
(b) To calculate the molar concentration of OH- in the 50.0 mL sample of the methylamine solution, we need to use the equation:
\[K_b = \frac{[\text{CH}_3\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{NH}_2]}\]
We have the value for Kb, which is 4.4 x 10^-4, and the concentration of CH3NH2 (aq) is unknown. Let's assume it to be x. At equilibrium, the concentration of CH3NH3+ and OH- is also x. Plugging in these values, we can solve for x:
\[4.4 \times 10^{-4} = \frac{(x)(x)}{(x)}\]
Simplifying the equation, we get:
\[4.4 \times 10^{-4} = x\]
Therefore, the molar concentration of OH- in the 50.0 mL sample of the methylamine solution is 4.4 x 10^-4 M.
(c) To calculate the initial molar concentration of CH3NH2 (aq) before it reacted with water and equilibrium was established, we need to understand the reaction stoichiometry. From the equation provided, we can see that for every mole of CH3NH2 (aq), one mole of CH3NH3+ is produced. Since the volume of the solution is 50.0 mL, we can convert it to liters by dividing by 1000:
\[50.0 \text{ mL} = 50.0 \times 10^{-3} \text{ L}\]
Now, using the concentration of CH3NH2 (aq) as x, we can set up the equation:
\[x = \frac{\text{moles of CH}_3\text{NH}_2}{\text{volume of solution in liters}}\]
Plugging in the values, we get:
\[x = \frac{x}{50.0 \times 10^{-3}}\]
Solving for x, we find:
\[x = 0.022 \text{ M}\]
Therefore, the initial molar concentration of CH3NH2(aq) in the solution before it reacted with water and equilibrium was established is 0.022 M.
(d) The net ionic equation that represents the reaction taking place during the titration is:
\[CH_3NH_2 (aq) + H^+ (aq) \rightarrow CH_3NH_3^+ (aq)\]
(e) To calculate the concentration of the HCl solution used to titrate the methylamine, we need to use the following equation:
\[M_1V_1 = M_2V_2\]
Where M1 is the molar concentration of the HCl solution, V1 is the volume of the HCl solution used (36.0 mL = 36.0 x 10^-3 L), M2 is the molar concentration of the methylamine solution (0.022 M), and V2 is the volume of the methylamine solution (50.0 mL = 50.0 x 10^-3 L).
Plugging in the values, we get:
\[M_1(36.0 \times 10^{-3}) = (0.022)(50.0 \times 10^{-3})\]
Simplifying the equation, we find:
\[M_1 = \frac{(0.022)(50.0 \times 10^{-3})}{36.0 \times 10^{-3}}\]
Calculating the value, we find:
\[M_1 = 0.0306 \text{ M}\]
Therefore, the concentration of the HCl solution used to titrate the methylamine is 0.0306 M.
(f) Unfortunately, as a Clown Bot, I am unable to sketch graphs or provide visual information. I apologize for the inconvenience. But hey, if you want a clown car instead, I'm your bot!
(a) The expression for the equilibrium constant, Kb, for methylamine is:
Kb = [CH3NH3+][OH-] / [CH3NH2]
(b) To calculate the molar concentration of OH- in the 50.0 mL sample of the methylamine solution, we need to use the formula for Kb.
Kb = [CH3NH3+][OH-] / [CH3NH2]
We know that the molar concentration of CH3NH3+ is equal to the molar concentration of OH- at equilibrium, so we can substitute [OH-] for [CH3NH3+] in the equation:
Kb = [OH-][OH-] / [CH3NH2]
Since the molar concentration of OH- is the same as the molar concentration of CH3NH3+, we can rewrite the equation as:
Kb = [OH-]^2 / [CH3NH2]
Now, we can rearrange the equation to solve for [OH-]:
[OH-]^2 = Kb * [CH3NH2]
[OH-] = sqrt(Kb * [CH3NH2])
Given that Kb = 4.4 x 10^-4 and the volume of the sample is 50.0 mL (0.0500 L), we can substitute these values into the equation:
[OH-] = sqrt(4.4 x 10^-4 * [CH3NH2])
(c) To calculate the initial molar concentration of CH3NH2(aq) in the solution before it reacted with water and equilibrium was established, we need to use the balanced equation and the initial volume of the sample.
From the balanced equation, we can see that the ratio of CH3NH2 to CH3NH3+ is 1:1. This means that the initial molar concentration of CH3NH2 is the same as the molar concentration of CH3NH3+, which we can find using the given pH.
pOH = 14 - pH
pOH = 14 - 11.77
pOH = 2.23
Since [OH-] = 10^(-pOH), we can calculate the molar concentration of OH-:
[OH-] = 10^(-2.23)
Now we can substitute this value into the equation from part (b) to find the initial molar concentration of CH3NH2:
[OH-] = sqrt(Kb * [CH3NH2])
sqrt(Kb * [CH3NH2]) = 10^(-2.23)
Kb * [CH3NH2] = (10^(-2.23))^2
[CH3NH2] = (10^(-2.23))^2 / Kb
(d) The net-ionic equation that represents the reaction that takes place during the titration is:
CH3NH2(aq) + HCl(aq) → CH3NH3+(aq) + Cl-(aq)
(e) To calculate the concentration of the HCl solution used to titrate the methylamine, we need to use the balanced equation and the volume of HCl solution added at the equivalence point.
From the balanced equation, we can see that the ratio of CH3NH2 to HCl is 1:1. This means that the moles of CH3NH2 reacted is equal to the moles of HCl added at the equivalence point.
Using the volume of HCl solution added at the equivalence point (36.0 mL or 0.0360 L) and the known molar concentration of HCl (which we need to calculate), we can calculate the moles of HCl:
moles HCl = molar concentration of HCl * volume of HCl solution added
Since moles of CH3NH2 reacted is equal to moles of HCl, we can set up the equation:
[CH3NH2] * initial volume of CH3NH2 solution = molar concentration of HCl * volume of HCl solution added
Substituting the values we have:
[CH3NH2] * 50.0 mL = molar concentration of HCl * 36.0 mL
Solving for the molar concentration of HCl:
molar concentration of HCl = ([CH3NH2] * 50.0 mL) / 36.0 mL
(f) Unfortunately, as text-based AI, I am unable to provide a graph. However, I can tell you that the titration curve will have a sharp increase in pH initially, followed by a more gradual increase as the equivalence point is approached. At the equivalence point, the pH will be equal to 7 (neutral), and then it will start decreasing as more HCl is added. The equivalence point should be labeled on the graph as the point where the pH reaches 7.
(a) The expression for the equilibrium constant, Kb, for methylamine can be obtained by writing the balanced chemical equation for the reaction and then expressing the concentration of the products over the concentration of the reactants.
The balanced chemical equation is: CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq)
The equilibrium expression, Kb, is written as: [CH3NH3+][OH-] / [CH3NH2]
(b) To calculate the molar concentration of OH- in the 50.0 mL sample of the methylamine solution, we need to first use the given pH of 11.77 to determine the concentration of H+.
pH is defined as the negative logarithm of the H+ concentration: pH = -log [H+]
Therefore, [H+] = 10^(-pH) = 10^(-11.77) mol/L
Since water dissociates to give equal concentrations of H+ and OH-, the concentration of OH- in the solution is also 10^(-11.77) mol/L.
(c) To calculate the initial molar concentration of CH3NH2(aq) in the solution before it reacted with water and equilibrium was established, we can use the fact that at the equivalence point of the titration, the number of moles of CH3NH2(aq) reacted is equal to the number of moles of H+ that were added.
Given that the volume of HCl solution added at the equivalence point is 36.0 mL, and assuming the concentration of HCl is C (unknown), the number of moles of H+ is (C x 0.036) mol.
Since CH3NH2 and H+ react in a 1:1 ratio, the initial molar concentration of CH3NH2(aq) is also equal to (C x 0.036) mol/L.
(d) The net-ionic equation that represents the reaction that takes place during the titration is:
CH3NH2(aq) + H+(aq) → CH3NH3+(aq)
(e) To calculate the concentration of the HCl solution used to titrate the methylamine, we need to use the volume and concentration information at the equivalence point.
Given that the volume of HCl solution added at the equivalence point is 36.0 mL, and assuming the initial concentration of HCl is C, we can set up the equation:
(C x 0.036) mol/L = (10^(-5.98)) mol/L
Solving for C, we find:
C = (10^(-5.98)) mol/L / 0.036
(f) Unfortunately, without the provided axes and graph, it is not possible for me to sketch the titration curve. However, I can describe it for you:
The titration curve will initially show a relatively flat portion, corresponding to the addition of the HCl solution, which does not significantly change the pH. At a certain point, called the equivalence point, a sudden change in pH will occur as the acid reacts completely with the base. This point is marked on the graph. After the equivalence point, the pH will decrease rapidly, indicating an excess of H+ ions in the solution.