Calculate the pH at the equivalence point for the titration of 0.210 M methylamine (CH3NH2) with 0.210 M HCl. The Kb of methylamine is 5.0×10−4.
5.84
Well, let's start by understanding what happens during the titration. Methylamine (CH3NH2) is a weak base, and HCl is a strong acid. When they react, they form the methylammonium ion (CH3NH3+) and chloride ion (Cl-).
During the titration, as HCl is added to methylamine, the acidic protons from HCl will react with the methylamine molecules until they are all consumed. This is known as the equivalence point.
At the equivalence point, all the methylamine has reacted with the HCl, forming methylammonium ions. Since methylammonium is the conjugate acid of methylamine, it can act as an acid and donate a proton. Therefore, the solution will be acidic.
To calculate the pH at the equivalence point, we need to find the concentration of the methylammonium ions. Since the initial concentration of both the methylamine and HCl is 0.210 M, at the equivalence point, the concentration of the methylammonium ions will also be 0.210 M.
Given that the Kb of methylamine is 5.0×10−4, we can write the equilibrium reaction as follows:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
Since Kb = [CH3NH3+][OH-] / [CH3NH2], we can rearrange the equation to solve for [OH-]:
[OH-] = (Kb * [CH3NH2]) / [CH3NH3+]
Substituting the values, we get:
[OH-] = (5.0×10−4 * 0.210) / 0.210
Simplifying further, we find [OH-] = 5.0×10−4 M.
Now, to calculate the pH, we need to find the pOH using the concentration of hydroxide ions:
pOH = -log[OH-] = -log(5.0×10−4) ≈ 3.3
Finally, we can find the pH using the equation:
pH = 14 - pOH = 14 - 3.3 ≈ 10.7
So, at the equivalence point, the pH will be around 10.7.
To calculate the pH at the equivalence point for the titration of methylamine (CH3NH2) with HCl, we need to consider the reaction that takes place during the titration.
Methylamine (CH3NH2) is a weak base, and HCl is a strong acid. The reaction between methylamine and HCl can be represented as follows:
CH3NH2 + HCl → CH3NH3+ + Cl-
At the equivalence point of the titration, an equal number of moles of methylamine and HCl have reacted. This means that the concentration of CH3NH2 will be reduced by the same amount as the concentration of HCl during the titration.
Since the concentration of HCl and methylamine is the same (0.210 M), at the equivalence point, the concentration of each will be halved (0.105 M).
Methylamine (CH3NH2) is a weak base, and its Kb value is given as 5.0×10^−4. The Kb expression for methylamine can be written as:
Kb = [CH3NH3+][OH-] / [CH3NH2]
Since we are interested in finding the pH, which is the negative logarithm of the concentration of H+, we need to calculate [H+].
At equilibrium, [CH3NH2] = [CH3NH3+] because the reaction is between a weak base and a strong acid.
Therefore, [CH3NH2] = 0.105 M and [CH3NH3+] = 0.105 M.
Now, let's calculate [OH-]:
Kb = [CH3NH3+][OH-] / [CH3NH2]
5.0×10^(-4) = (0.105 [OH-]) / 0.105
[OH-] = 5.0×10^(-4)
Since the solution is basic, [OH-] = 5.0×10^(-4) M, and we can use the pH equation to determine the pH:
pOH = -log[OH-]
pOH = -log(5.0×10^(-4))
pOH = -(-3.30)
pOH = 3.30
Since pH + pOH = 14 (at 25°C), we can find the pH:
pH = 14 - pOH
pH = 14 - 3.30
pH = 10.70
Therefore, the pH at the equivalence point for the titration of 0.210 M methylamine with 0.210 M HCl is approximately 10.70.
(CH3NH2) = (HCl) = 0.210 M; therefore, volumes will be equal which means at the equivalence point the concn of the salt will be 1/2 of 0.210 or 0.105 M.
CH3NH2 + HCl ==> CH3NH3Cl
..............CH3NH3^+ + HOH ==> CH3NH2 + H3O^+
I..............0.105................................0...............0
C..............-x....................................x................x
E..........0.105-x...............................x................x
Kb fo CH3NH3^+ = (Kw/Ka for CH3NH2) =
(x)(x)/(0.105-x). Solve for x = (H3O^+) and convert to pH.
Post your work if you get stuck.