Question

Calculate the pH of the solution that results from each of the following mixtures

A.) 150.0mL of 0.23M  HF with 225.0mL of 0.32M  NaF

B.) 185.0mL of 0.12M C2H5NH2 with 285.0mL of 0.22M C2H5NH3Cl

Answer 1

(a)
First we need the mols
0.150 L * 0.23 M = 0.0345 mols HF
0.225L * 0.32 M = 0.0720 mols F-

Now the total volume
0.150 + 0.225 = 0.375L

Then we can get the molarity of both
0.0345 mols / 0.375 = 0.092 M HF
0.0720 mols / 0.375 = 0.192 M F-

pKa of HF = 3.18
pH = 3.18 + log(0.192/0.092)
pH = 3.5

(b)
This is a buffer solution
always ask yourself what species are in the solution.
We are going to dissociate the salt first by converting the M to mol:

C₂H₅NH₃Cl ------> C₂H₅NH₃⁺ + Cl^-
0 ..........,,.............x moles.......x moles

Next we will dissociate the weak base.
Convert it to moles by multiplying the molarity with the liters first because you have to apply the volume given. Then make an ICE table for the base dissociation:

CH₃NH₂ + H₂O <---> CH₃NH₃⁺ + OH^-
y moles____.-.________x moles___y moles

Now convert them back into M by dividing the number of moles by the amount (L) of solution.
When you obtain the [OH-], follow what you did in part (a) to obtain the pH