Question

5.0 mL 1.0 M NaOAc + 28.0 mL 1.0 M HOAc

pH= 3.68

Calculations:

1. [H+] =?

2. [OAc-]=?

3. [HOAc]=?

4. Ka=?

Answer 1

Given that 5.0 mL 1.0 M NaOAc + 28.0 mL 1.0 M HOAc and pH = 3.68

Moles of NaOAc = molarity x volume in Litres = 1.0 M x 0.005 L = 0.005 moles

Moles of HOAc = molarity x volume in Litres = 1.0 M x 0.028 L = 0.028 moles

1) [H+] :

HOAc -----> H+ + OAc-

0.028 moles HOAc ----------------> 0.028 moles H⁺ + 0.028 moles OAc^-

[H+] = 0.028 moles

  [OAc-] = 0.028 moles

2) [OAc-]

NaOAc -----> Na+ + OAc-

0.005 moles NaOAc -----> 0.005 moles Na+ + 0.005 moles OAc-

[OAc-] = 0.005 moles

Total [OAc-] = 0.028 moles + 0.005 moles = 0.033 moles

3) [HOAc]:

moles of HOAc = 0.028

[HOAc] = 0.028 moles

4) Ka:

pH = pKa + log [NaOAc]/[HOAc]

pH = -logKa + log [NaOAc]/[HOAc]

3.68 = -logKa + log (0.005/0.028)

-logKa =   3.68 -log (0.005/0.028)

logKa = -[log (0.005/0.028) - 3.68]

Ka = 10 ^{-[ 3.68 -log (0.005/0.028)​]}

  Ka = 3.7 x 10^-5

Therefore, Ka = 3.7 x 10^-5