Calculation of Original pH from Final pH after Titration A biochemist has 100 mL of a 0.10 M solution of a weak acid with a pKa
2 answers:
Answer:
6.9
Explanation:
A weak acid dissociates in an equilibrium reaction, thus, it is in equilibrium with its conjugate base:
HA ⇄ H⁺ + A⁻
The equilibrium constant (Ka) can be calculated, where Ka = [H⁺]*[A⁻]/[HA]. Using the -log form, we can also have pKa = -logKa. By the Handerson-Halsebach (HH) equation, the relation between pH and pKa is:
pH = pKa + log[A⁻]/[HA]
So, when pH = 5.7, for this acid, the ratio of [acid]/[base] ([HA]/[A-]) is:
5.7 = 6.3 + log[A⁻]/[HA]
log[A⁻]/[HA] = -0.6
log[HA]/[A⁻] = 0.6
[HA]/[A⁻] =
[HA]/[A⁻] = 3.98 = 4.0
If the ratio of acid and base is 4 to 1, it means that 80%(4/5) of the acid is protonated after the addition of the HCl.
The initial number of moles of the weak acid was: 100 mL * 0.10 M = 10 mmol, so after the addition of HCl, 8 mmol is in the acid form (80% of 10). It was added 6.0 mmol of HCl (6.0 mL*1.0M). Thus, 6.0 mmol of H+ was added and reacted with the conjugate base of the weak acid.
For the mass conservation, the initial amount of the protonated weak acid must be 2.0 mmol (8 - 6), and the number of moles of the conjugate base was 8.0 mmol. Using the HH equation:
pH = 6.3 + log(8/2)
pH = 6.3 + 0.6
pH = 6.9
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