Answer:
[Na₂CO₃] = 0.094M
Explanation:
Based on the reaction:
HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)
It is possible to find pH using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻] / [HA]
Where [A⁻] is concentration of conjugate base, [CO₃²⁻] = [Na₂CO₃] and [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.
pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>
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Replacing these values:
10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]
<em> [Na₂CO₃] = 0.094M</em>
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Answer: umm, yes? DUH
Explanation: touch some falling out of a tap
Answer:
Explanation:
The charges are as follows:
since oxidation number of lead is 2 as indicated lead (II)nitrite. The number in parentheses indicates the oxidation number.
bears the charge -1.
As lead bears 2 positive charges so we need two nitrite ions to form lead (II) nitrite.
, it is a halogen and bears the charge -1.
, it is an alkali metal.
Since Bromide bears -1 charge and potassium bears +1 charge so we need only 1 ion of bromide and 1 ion of potassium to form potassium bromide.
Similarly, to form lead bromide we need 2 ions of bromide and 1 ion of lead, and to form potassium nitrite we need 1 nitrite ion and 1 potassium ion.
The final balanced equation is as follows, the various states of reactants and products are also written properly.
B would probably be the better choice but A is also partially correct I would go with (B) though.