I'm trying do this question:

0.10 mol/L hydrochloric acid is titrated with 0.10 mol/L methylamine, CH3NH2. Calculate the pH of the equivalence point.

I'm having trouble with the equation... How do you know what HCl + CH3NH2 will produce? Is there a rule or something for strong acid+weak base reactions? I know this won't produce water + a salt like usual neutralizations.
My guess - and it's really an uneducated guess - is that the equation will be like :
HCl + CH3NH2 -> CH3Cl + NH3 ?

I think I'll be able to figure out the rest of the question once I get the equation.

Thanks in advance!

To determine the reaction between hydrochloric acid (HCl) and methylamine (CH3NH2), you can apply the concept of acid-base reactions. Hydrochloric acid is a strong acid, while methylamine is a weak base. When a strong acid reacts with a weak base, the resulting products can be predicted using the concept of proton transfer.

In this case, HCl will donate a proton (H+) to CH3NH2, forming its conjugate acid, CH3NH3+ (methylammonium ion), and the chloride ion (Cl-):

HCl + CH3NH2 -> CH3NH3+ + Cl-

So you were on the right track with your guess! The chloride ion (Cl-) doesn't further react in this equation.

Now, let's move on to calculating the pH at the equivalence point. At the equivalence point, the moles of acid and base are equal, meaning the reaction has proceeded to completion. Since the amount of CH3NH2 is equal to HCl, we can consider them completely neutralized.

To determine the pH at the equivalence point, you need to identify the conjugate acid-base pair. In this case, the conjugate acid is CH3NH3+, and its associated base is methylamine, CH3NH2. Since CH3NH3+ is a weak acid, you can use the Kb (base dissociation constant) to calculate the pOH and subsequently convert it to pH.

The Kb for methylamine (CH3NH2) is usually provided in the question or can be looked up in a reference source. Let's say for this example, the Kb value is 4.38 x 10^-4.

Using the appropriate equilibrium expression for the Kb of CH3NH2 and assuming x is the concentration of OH- formed during the reaction, you can set up the equation:

Kb = [CH3NH3+][OH-] / [CH3NH2]

Since the concentration of OH- is equal to the concentration of CH3NH3+ at the equivalence point, you can substitute [OH-] = [CH3NH3+] into the equation:

Kb = [CH3NH3+]^2 / [CH3NH2]

[CH3NH3+] = (Kb * [CH3NH2])^(1/2)

Now, since [CH3NH3+] = [OH-], you can find the pOH by taking the -log of the concentration of OH-:

pOH = -log([OH-]) = -log([CH3NH3+])

Finally, calculate the pH by subtracting the pOH from 14:

pH = 14 - pOH

By plugging in the Kb value and the concentration of CH3NH2 (0.10 mol/L), you can calculate the pH at the equivalence point.