<span>A) This solution was not basic when it was heated in part 3. ( in part 3 i convertedCu(OH)2 to CuO).
Incorrectly low, because not all copper compounds will precipitate out
B) A slightly blue solution was decanted from Cu in part V. (in part 5 i reduced Cu(H20)6 ions with zink)
Incorrectly low, because some copper were thrown away
C) In part 5 the water in the beaker boiled away, exposing the evaporating dish to excess heat (same as above).
incorrectly high, because other compounds might be present as well </span>
Henderson-Hasselbalch equation relates the pH with the dissociation constant of the acid. The pH of the buffer solution will be 3.90.
<h3>What is the Henderson-Hasselbalch equation?</h3>
The Henderson-Hasselbalch equation is used to calculate the pH or the concentration of the conjugate base and acid.
The Henderson-Hasselbalch equation can be given as,
pH = pKa + log [A⁻] ÷ [HA]
The dissociation reaction is given as,
HA⁺ + H2O ⇌ H3O⁺ + A⁻
NaA → Na⁺ + A⁻
For this first pKa is calculated as:
pKa = - log (1. 0 x 10⁻⁴)
Substituting the value of pKa in Henderson-Hasselbalch equation pH is determined as:
pH = - log (1. 0 x 10⁻⁴) + log [0.08] ÷ [0.1]
= 4 + (-0.0969)
= 3.90
Therefore, 3.90 is the pH of the solution.
Learn more about Henderson-Hasselbalch here:
brainly.com/question/13151501
#SPJ4