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][H20]
b) -log[H+]=pH=11.77
pH+pOH=14.00
14.00-11.77=2.23=pOH
[OH-]=10^-pOH=10^(-2.23)=.0058M
Everything Christina did is correct.
Starting with C
4.4E-9=[OH-]^2/x x=.078
d. Ch3Nh2 + H+ -> CH3NH3+
e.(30)(x)=(50)(.078) since moles are equal at equivalence.
x=.13 Molar
in your equilibrium constant expression do not include water because it is a liquid
Why is the [] of the CH3NH3 the same as the [OH-]?
(a) The equilibrium constant expression for the reaction is determined by considering the products and reactants in the equation. In the given reaction, the equilibrium constant expression, Kb, can be written as:
Kb = [CH3NH3+][OH-]/[CH3NH2]
(b) To calculate the molar concentration of OH- in the 50.0 mL sample, you need to know the concentration of CH3NH3+ in equilibrium. However, it is not provided in the question. So, it is not possible to directly determine the molar concentration of OH-.
(c) To calculate the initial molar concentration of CH3NH2(aq) in the solution before it reacted with water and equilibrium was established, you need to use the given pH of the solution.
The pH of a solution can be calculated using the equation:
pH = -log[H+]
In this case, the pH is given as 11.77. To calculate the concentration of [H+], you can use the relationship:
[H+] = 10^(-pH)
[H+] = 10^(-11.77)
Once you have the concentration of [H+], you can assume that the initial concentration of CH3NH2(aq) is equal to [OH-] since they react in a 1:1 ratio. So, the initial molar concentration of CH3NH2(aq) is equal to [OH-].
(d) The net ionic equation for the titration reaction between CH3NH2(aq) and HCl(aq) can be written as:
CH3NH2(aq) + HCl(aq) → CH3NH3+(aq) + Cl-(aq)
(e) To calculate the concentration of the HCl solution used to titrate the methylamine, you need to use the stoichiometry of the balanced equation and the volume information.
From the balanced equation in part (d), you can see that the mole ratio of CH3NH2(aq) to HCl(aq) is 1:1. Therefore, the amount of moles of HCl used in the titration is equal to the amount of moles of CH3NH2 in the original solution.
Using the concentration of the CH3NH2(aq) calculated in part (c) and the volume information provided (36.0 mL of HCl solution), you can calculate the concentration of the HCl solution as follows:
Moles of CH3NH2(aq) = initial concentration of CH3NH2(aq) * volume of CH3NH2(aq) solution (in L)
Moles of HCl = initial concentration of HCl * volume of HCl solution (in L)
Since the moles of CH3NH2(aq) and HCl are equal, you can calculate the concentration of the HCl solution:
Concentration of HCl = Moles of HCl / Volume of HCl solution (in L)
(f) Unfortunately, as a text-based AI, I am unable to provide a visual representation such as a graph. However, I can describe the titration curve for you.
The titration curve starts with a slightly basic solution of CH3NH2(aq) before titration. As HCl solution is added, the CH3NH2(aq) is neutralized, leading to a gradual decrease in pH. Near the equivalence point (36.0 mL of HCl solution), the pH reaches its lowest value, which is 5.98 according to the question. After the equivalence point, the excess HCl converts the remaining CH3NH2(aq) to CH3NH3+(aq), resulting in a sharp increase in pH.
The equivalence point of the titration corresponds to the point where the stoichiometric amount of acid has reacted with the base. In this case, it is reached when 36.0 mL of HCl solution has been added.