Answer:
4.27
Explanation:
Let's consider the dissociation of a generic monoprotic acid.
HA(aq) → H⁺(aq) + A⁻(aq)
The pH is 2.36. The concentration of H⁺ is:
pH = -log [H⁺]
[H⁺] = antilog -pH
[H⁺] = antilog -2.36 = 4.37 × 10⁻³ M
We know that the concentration of the acid Ca = 0.3535 M. We can find the acid dissociation constant using the following expression.
[H⁺] = √(Ca × Ka)
Ka = [H⁺]²/Ca
Ka = (4.37 × 10⁻³)²/0.3535
Ka = 5.40 × 10⁻⁵
The pKa is:
pKa = -log Ka = -log 5.40 × 10⁻⁵ = 4.27
Answer:
25th electron
Explanation:
the last electron is the valence electron. Assuming it has equal numbers of protons and electrons, then the 25th electron is the valence.
Answer:
At the second equivalent point 200 mL of NaOH is required.
Explanation:
at the first equivalent point:
H2A + OH- = HA- + H2O
initial mmoles y*100 y*100 - -
final mmoles 0 0 y*100 y*100
at the second equivalent point:
HA- + OH- = A2- + H2O
initial mmoles y*100 y*100 - -
final mmoles - - y*100 y*100
at the second equivalent point we have that y*100 mmoles of NaOH or 100 mL of NaOH ir required, thus:
at the second equivalent point 200 mL of NaOH is required.
50.00ml*(10^-3L/ml)*(3.91moles/L) = 0.196 mol
There are two electrons in 4s and 6 in 3d. When you lose 3 electrons, you will be left with an ion of 3+ charge!
