Methyl amine is a weak base with a pKb=3.35. Consider the titration of 30.0mL of .030M of Methyl amine with 0.025M HCl.

a) Write the appropriate equation for the reaction.
b) Calculate K for the reaction in part (a).
c) Calculate pH of the initial methyl amine solution (0mL added HCl)
d) calculate pH of solution with addition of 10mL HCL

CHNH2 + HCl ==> CH3NH3Cl

b. Look up the delta Go values in your text/notes and solve for dGo for the rxn, then dGo = -RTlnK. Substitute and solve for K.

c.
..CH3NH2 + HOH ==> CH3NH3^+ + OH^-
I..0.030............0..........0
C.....-x............x..........x
E..0.030-x..........x..........x

Convert pKb to Kb (pKb = -log Kb). Substitute the E line into the Kb expression and solve for x = OH^- and convert to H^+ then to pH.

d.
millimols CH3NH2 initially = mL x M = 30 mL x 0.030 M = 0.9
Therefore, CH3NH3Cl formed is 10 mL x 0.025 = 0.25 mmols and the CHNH2 used is 0.25 which leaves 0.9-0.25 = 0.65. Substitute the 0.65 and the 0.25 into the Henderson-Hasselbalch equation and solve for pH. You will need pKa for that and you get it from pKa + pKb = pKw = 14. With pKw and pKb, you can solve for pKa.

a) The appropriate equation for the reaction is:

CH3NH2 + HCl -> CH3NH3+Cl-

b) To calculate K, we need to consider the equilibrium reaction of the base, methyl amine, with the acid, HCl:

CH3NH2 + HCl <-> CH3NH3+Cl-

The equilibrium constant K can be calculated using the concentrations (or molarities) of the products and reactants:

K = [CH3NH3+][Cl-] / [CH3NH2][HCl]

However, we're given the pKb value of methyl amine, which is a measure of the strength of the base. The relationship between pKb and Kb (equilibrium constant for base strength) is:

pKb = -log10(Kb)

From this, we can calculate Kb:

Kb = 10^(-pKb)

c) To calculate the pH of the initial methyl amine solution (0 mL added HCl), we need to determine the concentration of the methyl amine in the solution. We know the volume of methyl amine solution and its concentration:

Volume of methyl amine solution (V1) = 30.0 mL
Concentration of methyl amine (C1) = 0.030 M

To calculate the moles of methyl amine (n1) in the solution, we use the formula:

n1 = C1 * V1

Now, using the equation for Kb, we can calculate the OH- concentration ([OH-]):

Kb = [OH-]^2 / [CH3NH2]

[OH-] = sqrt(Kb * [CH3NH2])

Since [OH-] = [CH3NH3+], we can use the equation for pOH to calculate the pOH of the solution:

pOH = -log10([OH-])

Then, to find the pH, we can use the relationship:

pH + pOH = 14

d) To calculate the pH of the solution with the addition of 10 mL of HCl, we need to consider the reaction between methyl amine and HCl. The equation for this reaction can be written as:

CH3NH2 + HCl -> CH3NH3+Cl-

We know the volume of HCl added (V2) is 10 mL and its concentration (C2) is 0.025 M. The total volume of the solution after the addition of HCl is 30.0 mL + 10 mL = 40.0 mL.

To calculate the new concentration of methyl amine (C1'), in the 40.0 mL solution, we use the formula:

C1' * V1' = C1 * V1

Where:
V1' = V1 (initial volume of methyl amine) = 30.0 mL
C1' = concentration of methyl amine after the addition of HCl

Once we have the new concentration of methyl amine, we can follow the same steps as in part c) to calculate the pH of the solution.