Answer:
pKb = 10.96
Explanation:
Tartaric acid is a dyprotic acid. It reacts to water like this:
H₂Tart + H₂O ⇄ H₃O⁺ + HTart⁻ Ka1
HTart⁻ + H₂O ⇄ H₃O⁺ + Tart⁻² Ka2
When we anaylse the base, we have
Tart⁻² + H₂O ⇄ OH⁻ + HTart⁻ Kb1
HTart⁻ + H₂O ⇄ OH⁻ + H₂Tart Kb2
Remember that Ka1 . Kb2 = Kw, plus pKa1 + pKb2 = 14
Kb2 = Kw / Ka1 → 1×10⁻¹⁴ / 9.20×10⁻⁴ = 1.08×10⁻¹¹
so pKb = - log Kb2 → - log 1.08×10⁻¹¹ = 10.96
__ KClO₃ → __ KCl + __ O₂
Left Side:
1 K
1 Cl
3 O
Right Side:
1 K
1 Cl
2 O
Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.
This gives us 2KClO₃ → __ KCl + 3O₂.
However, this equation is still not balanced.
Left Side:
2 K
2 Cl
6 O
Right Side:
1 K
1 Cl
6 O
In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.
2KClO₃ → 2KCl + 3O₂
False. The number of neutrons can be more or less than the number of protons.
