HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)
NH4Cl(aq) + +NaOH(aq) -> NH3(aq) + NaCl(aq) + H2O(l)
NH3(aq) + HCl(aq) -> NH4Cl(aq)
Write net ionic equations for the three reactions. Show how the first two reactions are arranged algebraically to determine the third.
1. H^+ + OH^- ==> H2O
2. NH4^+ + OH^- ==> NH3 + H2O
3. NH3 + H^+ ==> NH4^+
Add equation 1 to rhe reverse of equation 2 to obtain equation 3.
You should add the phase to each
Sure! Let's write the net ionic equations for each reaction first:
1. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Net ionic equation: H+(aq) + OH-(aq) → H2O(l)
2. NH4Cl(aq) + NaOH(aq) → NH3(aq) + NaCl(aq) + H2O(l)
Net ionic equation: NH4+(aq) + OH-(aq) → NH3(aq) + H2O(l)
Now, let's arrange the first two reactions algebraically to determine the third:
1. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Net ionic equation: H+(aq) + OH-(aq) → H2O(l)
2. NH4Cl(aq) + NaOH(aq) → NH3(aq) + NaCl(aq) + H2O(l)
Net ionic equation: NH4+(aq) + OH-(aq) → NH3(aq) + H2O(l)
By comparing the net ionic equations of the first two reactions, we can see that both reactions involve the formation of water molecule (H2O) from H+ ion and OH- ion. This means that the OH- ion from the second reaction can be derived from the H2O molecule in the first reaction by losing one H+ ion.
When we combine the two net ionic equations, we can see that the H2O molecule appears on both sides of the equation. So, we can cancel it out, yielding the net ionic equation for the third reaction:
NH4+(aq) + H+(aq) → NH4Cl(aq)
Therefore, the net ionic equation for the third reaction is: NH4+(aq) + H+(aq) → NH4Cl(aq).
To write the net ionic equations for the three reactions, we first need to identify the species that are present in the reaction and determine their charges.
1) HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)
In this reaction, the complete ionic equation can be written as:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + Cl-(aq) + H2O(l)
To obtain the net ionic equation, we eliminate the spectator ions (ions that appear on both sides of the equation, in this case Na+ and Cl-), leaving only the ions involved in the chemical change. The net ionic equation is:
H+(aq) + OH-(aq) -> H2O(l)
2) NH4Cl(aq) + NaOH(aq) -> NH3(aq) + NaCl(aq) + H2O(l)
The complete ionic equation for this reaction can be written as:
NH4+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) -> NH3(aq) + Na+(aq) + Cl-(aq) + H2O(l)
Using the same process, we eliminate spectator ions to obtain the net ionic equation:
NH4+(aq) + OH-(aq) -> NH3(aq) + H2O(l)
3) To determine the third reaction (NH3(aq) + HCl(aq) -> NH4Cl(aq)), we can consider the first two reactions algebraically.
By comparing reactions 1 and 2, we can see that OH- and HCl (H+) are common species in both reactions. To get the net ionic equation for the third reaction, we can eliminate these common ions and focus on the remaining ions:
From reaction 1: H+(aq) + OH-(aq) -> H2O(l)
From reaction 2: NH4+(aq) + OH-(aq) -> NH3(aq) + H2O(l)
By adding these two equations together, we get:
H+(aq) + NH4+(aq) -> NH3(aq) + H2O(l)
Taking into account the charges of the ions, we can rewrite this equation as:
NH3(aq) + HCl(aq) -> NH4Cl(aq)
So, the net ionic equation for the third reaction is NH3(aq) + HCl(aq) -> NH4Cl(aq).