For the acid-base neutralization reaction of Carbonic acid, H2CO3(aq) and lithium hydroxide, LiOH(aq), write the blanced:
a)complete equation
b) Ionic equation
c) Net ionic equation.
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i got
a) H2CO3(aq) + 2LiOH(aq) = 2H2O(l) +Li2CO3(aq)
i know my answers for the next are not right i'm just confused how to start these equations.Please help! :)
To write net ionic equations you follow some rules. Some things that are written as molecules are
1. weak electrolytes (those that do not ionize 100% such as weak acids(those with a Ka), weak bases (those with a Kb), H2O, ETC.
b. insoluble materials (memorize the solubility rules). Here is a simplified set of rules. http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
c.gases
Everything else is written as ions.
H2CO3(aq) + 2Li^+(aq) + 2OH^-(aq) ==>2H2O(l) + 2Li^+(aq) + CO3^2-(aq)
What I have written above is the ionic equation. To write the net ionic equation, we cancel any ions common to both sides. Therefore, we cancel 2Li^+(aq) and rewrite the net ionic equation as
H2CO3(aq) + 2OH^-(aq) ==>2H2O(l) + CO3^2-(aq)
HI(aq)+Al(s)→
Express your answer as a balanced chemical equation. Identify all of the phases in your answer.
and
H2SO4(aq)+TiO2(s)→
Express your answer as a balanced chemical equation. Identify all of the phases in your answer.
along with:
Write a balanced chemical equation showing how each metal oxide reacts with HCl.
SrO
Na2O
Li2O
BaO
K
Ca
Sr
No problem, I can help you with that!
To write the ionic and net ionic equations, let's first start by determining the dissociation of each compound in water:
Carbonic acid, H2CO3, is a weak acid and it partially dissociates in water:
H2CO3(aq) ⇌ H+(aq) + HCO3^-(aq)
Lithium hydroxide, LiOH, is a strong base and it fully dissociates in water:
LiOH(aq) → Li+(aq) + OH^-(aq)
a) Complete equation:
To write the complete equation, we can simply combine the dissociated ions from both the acid and the base:
H2CO3(aq) + 2LiOH(aq) → 2H2O(l) + Li2CO3(aq)
Your answer for the complete equation is correct.
b) Ionic equation:
In the ionic equation, we separate the compounds into their respective ions:
H2CO3(aq) + 2LiOH(aq) → 2H2O(l) + Li2CO3(aq)
However, Li2CO3 is a strong electrolyte and will dissociate into ions as well. So the ionic equation would be:
2H+(aq) + CO3^2-(aq) + 2Li+(aq) + 2OH^-(aq) → 2H2O(l) + 2Li+(aq) + CO3^2-(aq)
c) Net ionic equation:
Net ionic equation shows only the species that undergo a change during the reaction, excluding spectator ions that appear on both sides of the equation.
In this case, the H+(aq) and OH^-(aq) ions combine to form water:
2H+(aq) + 2OH^-(aq) → 2H2O(l)
This is the net ionic equation for the acid-base neutralization reaction between carbonic acid and lithium hydroxide.
I hope this helps! Let me know if you have any more questions.
No worries! I'm here to help you understand how to balance the equations correctly.
a) To write the complete equation, you've correctly identified the reactants and products. Here's the balanced equation for the complete reaction:
H2CO3(aq) + 2LiOH(aq) → 2H2O(l) + Li2CO3(aq)
b) Now let's move on to writing the ionic equation. In the ionic equation, we separate all the soluble ionic compounds into their respective ions. Water (H2O) is a covalent compound and remains unchanged in its molecular form, so it does not split into ions.
H2CO3(aq) + 2LiOH(aq) → 2H2O(l) + Li2CO3(aq)
The ions in this equation are:
H2CO3 → 2H+ + CO3^2-
LiOH → Li+ + OH-
So, the ionic equation becomes:
2H+ + CO3^2- + 2Li+ + 2OH- → 2H2O(l) + Li2CO3(aq)
c) The net ionic equation represents only the species that participate in the reaction; spectator ions, which do not undergo any change, are omitted. In this case, the spectator ions are the Li+ and OH- ions on both sides of the equation.
The net ionic equation is obtained by eliminating the spectator ions from the ionic equation:
2H+ + CO3^2- → 2H2O(l) + Li2CO3(aq)
So, the net ionic equation for the acid-base neutralization reaction of carbonic acid (H2CO3) and lithium hydroxide (LiOH) is:
2H+ + CO3^2- → 2H2O(l) + Li2CO3(aq)
Remember, the net ionic equation focuses only on the species actively involved in the reaction, excluding spectator ions.