Calculate the pH at the equivalence point in the titration of 50ml of 0.20 M methylamine (Kb=4.3*10^-4) with a 0.40 M HCL solution.
To find the pH at the equivalence point, we first need to determine the moles of methylamine and hydrochloric acid that react during the titration.
Step 1: Calculate the moles of methylamine:
Moles of methylamine = volume (in liters) x molarity
Moles of methylamine = 0.050 L x 0.20 mol/L
Moles of methylamine = 0.010 mol
Step 2: Calculate the moles of hydrochloric acid:
Since hydrochloric acid is a strong acid, it will completely dissociate in water. So, the moles of hydrochloric acid will be equal to the volume (in liters) multiplied by the molarity.
Moles of hydrochloric acid = 0.050 L x 0.40 mol/L
Moles of hydrochloric acid = 0.020 mol
Step 3: Determine the limiting reactant:
The limiting reactant is the one with the smallest number of moles, which is methylamine in this case.
Step 4: Calculate the excess moles of the limiting reactant:
Excess moles of methylamine = Moles of methylamine - Moles of hydrochloric acid
Excess moles of methylamine = 0.010 mol - 0.020 mol
Excess moles of methylamine = -0.010 mol (negative value indicates excess)
At the equivalence point, the excess moles of methylamine will react with water to form the conjugate acid of methylamine, which is methylammonium ion (CH3NH3+).
Step 5: Calculate the concentration of the methylammonium ion:
The concentration of the methylammonium ion can be determined using the excess moles and the volume of the solution at the equivalence point.
Volume of the solution = Volume of methylamine + Volume of hydrochloric acid
Volume of the solution = 0.050 L + 0.050 L
Volume of the solution = 0.100 L
Concentration of methylammonium ion = Excess moles / Volume of the solution
Concentration of methylammonium ion = (-0.010 mol) / (0.100 L)
Concentration of methylammonium ion = 0.10 mol/L
Step 6: Calculate the pH:
To find the pH, we need to determine the concentration of hydroxide ions (OH-) resulting from the hydrolysis of methylammonium ion.
Kb = [OH-][CH3NH3+] / [CH3NH2]
Since the concentration of OH- is not given, we'll assume that it is small compared to the initial concentration of methylamine, allowing us to make the approximation that [CH3NH2] = [CH3NH3+] at the equivalence point.
Kb = [OH-][CH3NH3+] / [CH3NH2]
Kb = [OH-][CH3NH3+] / [CH3NH3+]
Kb = [OH-]
pOH = -log10(Kb)
pOH = -log10(4.3 x 10^-4)
pOH = 3.37
pH = 14 - pOH
pH = 14 - 3.37
pH = 10.63
Therefore, the pH at the equivalence point in the titration is approximately 10.63.
To calculate the pH at the equivalence point in the titration of methylamine (CH3NH2) with HCl, we need to determine the number of moles of each species present at the equivalence point, and then calculate the resulting pH.
Step 1: Write the balanced equation for the reaction between methylamine and HCl:
CH3NH2 + HCl -> CH3NH3+ + Cl-
It is a 1:1 ratio reaction. So, the number of moles of methylamine (CH3NH2) will be equal to the number of moles of hydrochloric acid (HCl) at the equivalence point.
Step 2: Calculate the number of moles of methylamine present in 50 ml of 0.20 M solution:
Moles of CH3NH2 = volume (L) x concentration (mol/L)
= 0.050 L x 0.20 mol/L
= 0.010 mol
Step 3: Calculate the number of moles of hydrochloric acid required to react with methylamine:
Moles of HCl = moles of CH3NH2 = 0.010 mol
Step 4: Calculate the volume of hydrochloric acid solution required to react with methylamine:
Volume of HCl = moles of HCl / concentration of HCl
= 0.010 mol / 0.40 mol/L
= 0.025 L = 25 ml
Step 5: Calculate the final volume of the combined solution:
The original volume of methylamine solution was 50 ml.
We added 25 ml of HCl solution.
Therefore, the final volume of the combined solution is 50 ml + 25 ml = 75 ml.
Step 6: Calculate the concentration of the resulting solution:
Total moles = moles of methylamine + moles of HCl
= 0.010 mol + 0.010 mol
= 0.020 mol
The final concentration is moles / volume
= 0.020 mol / 0.075 L
= 0.27 M
Step 7: Determine the pOH at the equivalence point:
Since methylamine is a weak base, we can use the Kb value to determine the pOH at the equivalence point.
Kb = [CH3NH3+][OH-] / [CH3NH2]
Since the solution is dilute, the [OH-] is approximately equal to the concentration of methylamine.
So, Kb = [OH-]^2 / [CH3NH2]
Let x be the concentration of [OH-] at the equivalence point, then:
Kb = x^2 / 0.27
Solving for x:
x = √(Kb * 0.27)
= √(4.3 * 10^-4 * 0.27)
= 0.012
Step 8: Calculate the pOH at the equivalence point:
Using the equation pOH = -log[OH-], we can find the pOH.
pOH = -log(0.012)
= 1.92
Step 9: Calculate the pH at the equivalence point:
Since pH + pOH = 14, we can find the pH.
pH = 14 - pOH
= 14 - 1.92
= 12.08
Therefore, the pH at the equivalence point in the titration is approximately 12.08.
How many mL HCl is needed? That's
50 x 0.2/0.4 = 25.00 mL
Total volume = 75.00 mL
Total mols = 0.050 x 0.2 =0.01
(salt at equivalence point) = 0.01/0.075L = approximately 0.13
If we call methylamine just BNH2 then the salt will be BNH3^+ and it will hydrolyze as follows:
......BNH3^+ + H2O ==> H3O^+ + BNH2
I.....0.13..............0........0
C.......-x..............x........x
E...-0.13-x.............x........x
Ka for BNH3^+ = (Kw/Kb for BNH2) = (x)(x)/(0.13-x)
Solve for x = (H3O^+) and convert to pH.