What is the net ionic equation when H2SO4 is neutralized by NaOH in aqueous solution?
Write the molecular equation.
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
Convert to ions.
2H^+ + SO4^= + 2Na^+ + 2 OH^- ==> 2Na^+ + SO4^= + 2H2O
Now cancel those ions that appear on both sides of the equation. What is left is the net ionic equation.
I shall be happy to check your answer.
2H^+ + 2OH^- --> 2H20
Thanks
for asking.
You're welcome! Your net ionic equation is correct:
2H+ + 2OH- --> 2H2O
This equation shows the balanced reaction between the hydrogen ions (H+) from sulfuric acid (H2SO4) and the hydroxide ions (OH-) from sodium hydroxide (NaOH), resulting in the formation of water (H2O).
The net ionic equation is obtained by canceling out the spectator ions that appear on both sides of the equation. In this case, the spectator ion is the sulfate ion (SO4^=), which does not participate in the actual reaction and remains unchanged. Therefore, it is not included in the net ionic equation.
If you have any more questions or need further clarification, feel free to ask!
Your net ionic equation is correct:
2H^+ + 2OH^- --> 2H2O
This equation represents the reaction of H2SO4 (sulfuric acid) being neutralized by NaOH (sodium hydroxide) in aqueous solution. The sulfate ion (SO4^=) and sodium ion (Na^+) are spectator ions and do not participate in the reaction. The two hydrogen ions (H^+) from the acid react with the two hydroxide ions (OH^-) from the base to form water (H2O).