Question

Use the Henderson-Hasselbalch equation to calculate the pH of each solution:

1)a solution that is 0.170 M in HC2H3O2 and 0.115 M in KC2H3O2

2) a solution that is 0.230 M in CH3NH2 and 0.130 M in CH3NH3Br

Answer 1

For your reference,

Henderson-Hasselbalch equation is :

pH = pK_a + log([Proton Acceptor]/[Proton Donor])

or

pH = pK_a + log([A^-] / [HA])

Now,

Solution 1 :

CH₃COOK gives CH₃COO^- which is A^- in this case.

and CH₃COOH is HA.

[A^-] = 0.115 M

[HA] = 0.170 M

NOTE: pK_a should either be given in the question or K_a should be given so that you can find pK_a by : pK_a = -log(K_a)

pK_a for acetic acid is 4.74

Now,

pH = pK_a + log([A^-] / [HA])

pH = 4.74 + log(0.115/0.170) = 4.74 - 0.17 = 4.57

Solution 2 :

[A^-] = [CH₃NH₃Br] = 0.130 M

[HA] = [CH₃NH₂] = 0.230 M

NOTE: pK_a should either be given in the question or K_a should be given so that you can find pK_a by : pK_a = -log(K_a)

pK_a for CH₃NH₂ is 3.36

Now,

pH = pK_a + log([A^-] / [HA])

pH = 3.36 + log(0.130/0.230) = 3.36 - 0.25 = 3.11

Answer might be a little different since it might have been calculated with an approximated value of pK_a

Thanks. Comment if any doubt.