Calculate the pH of a mixture of 25 cm3 of 0.100 mol dm–3 propanoic acid and 15 cm3 of 0.050 mol dm–3 sodium hydroxide. (pKa propanoic acid = 4.87)

Answer 1

Use the Henderson-Hasselbach equation: pH= pK_a + log/[base]/[acid]. the acid is propanoic and there are 0.0025 moles of it (Volume)(M). 1 dm³ = 1 liter.So (0.025)(0.1)=0.0025moles acid. For the base (0.015L)(0.05M)=7.5 X 10^-4, or 0.00075 moles. So 0.00075/0.0025=0.3. pK_a = 4.87 + log 0.3=4.34

The equation for the reaction is CH₃CH₂COOH + NaOH---> CH₃CH₂COO^-Na⁺ + H₂O.

Answer 2

When NaOH is added to the weak acid 3-propanoic acid, one forms a buffer because you have now created some conjugate base. You then have a mixture of weak acid and the salt of that weak acid. This is a buffer.

If HA represents the weak acid, then you have the following:

HA + OH- ==> A- + H2O where A- is the salt of the acid (the conjugate base)

moles HA: 0.025 L x 0.100 mol/L = 0.0025 moles

moles OH-: 0.015 L x 0.050 mol/L = 0.00075 moles

Set up an ICE table:

HA + OH- ==> A- + H2O

.0025... .00075.........0................Initial

-00075..-.00075...+.00075...........Change

0.00175......0...........0.00075.......Equilibrium

Use Henderson Hasselbalch: pH = pKa + log[salt]/[acid]

pH = 4.87 + log (0.00075/0.00175)

pH = 4.87 + log 0.429

pH = 4.87 + (-0.37)

pH = 4.50