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Brums [2.3K]
3 years ago
11

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

of 6.3. She adds 6.0 mL of 1.0 M HCl, which changes the pH to 5.7. What was the pH of the original solution?
Chemistry
2 answers:
ddd [48]3 years ago
8 0

Answer:

!

Explanation:

need it too^^

KATRIN_1 [288]3 years ago
3 0

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⁻] = 10^{0.6}

[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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