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?

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

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying

the correct answer is:

a. formation of precipitation