Calculate the pH at the equivalence point in the titration of 55.0 mL of 0.180 M methylamine(Kb = 4.4 × 10−4) with 0.330 M HCl.
To calculate the pH at the equivalence point of a titration, you need to determine the amount of acid and base that react completely, and then use this information to find the concentration of the resulting solution.
In this titration, you are given the volume (55.0 mL) and concentration (0.180 M) of methylamine, as well as the concentration (0.330 M) of HCl. The first step is to calculate the number of moles of each reactant.
Moles of methylamine = volume x concentration = 55.0 mL x 0.180 M = 9.9 mmol
Moles of HCl = volume x concentration = 55.0 mL x 0.330 M = 18.15 mmol
Since the balanced chemical equation for the reaction between methylamine and HCl is:
CH3NH2 + HCl → CH3NH3+ + Cl-
You can see that 1 mole of methylamine reacts with 1 mole of HCl. Therefore, the reactants are in a 1:1 ratio.
At the equivalence point, all the methylamine will react with the HCl, meaning that the limiting reactant will be either methylamine or HCl, depending on which is present in a smaller amount. In this case, methylamine is the limiting reactant because it has fewer moles.
After the reaction, the methylamine is completely converted into its conjugate acid, CH3NH3+. Since the volume remains the same (55.0 mL), the concentration of the resulting CH3NH3+ solution can be calculated by dividing the number of moles of CH3NH3+ by the volume (in liters):
Concentration of CH3NH3+ = moles / volume = 9.9 mmol / 55.0 mL = 0.18 M
The concentration of CH3NH3+ is identical to the concentration of Cl-, as both are formed in a 1:1 ratio. Therefore, the concentration of Cl- is also 0.18 M.
Now that you have the concentration of CH3NH3+ (the conjugate acid), you can calculate the pOH at the equivalence point using the Kb value for methylamine.
Kb = [CH3NH3+][OH-] / [CH3NH2]
Since the concentration of CH3NH3+ and CH3NH2 are identical at the equivalence point, you can square the concentration to simplify the equation:
Kb = [OH-]² / [CH3NH2]
Plugging in the known values:
4.4 × 10^−4 = [OH-]² / 0.18
To solve for [OH-], rearrange the equation:
[OH-]² = 4.4 × 10^−4 * 0.18
[OH-]² = 7.92 × 10^−5
Now, take the square root of both sides:
[OH-] = √(7.92 × 10^−5)
[OH-] = 0.0089 M
Since pH = 14 - pOH, you can calculate the pH at the equivalence point:
pOH = -log([OH-]) = -log(0.0089) = 2.05
pH = 14 - 2.05 = 11.95
Therefore, the pH at the equivalence point in the titration is approximately 11.95.