What is the net ionic equation for the reaction between Ba(OH)2 and H2SO4? [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 3 years ago.








[Improve this question](/posts/131213/edit) 



I know its either:


1.) Ba + SO4 = BaSO4


2.) Ba + 2OH + 2H + SO4 = BaSO4 + 2H2O


But can someone explain why water is or isn't part of the net ionic?




The reaction between $\ce{Ba(OH)2, H2SO4}$ is known as acid-base neutralisation, as $\ce{Ba(OH)2}$ is a relatively strong base and $\ce{H2SO4}$ the strong acid.


The molecular reaction is given as,


$$\ce{Ba(OH)2 + H2SO4 -> BaSO4 (s) + 2H2O}$$



> 
> Before moving to net-ionic reaction, let's see the reaction of $\ce{BaCl2, Na2SO4}$ **[1]**:
> 
> 
> $$\ce{BaCl2 + Na2SO4 -> BaSO4 (s) + 2NaCl}$$
> 
> 
> So, breaking up into ions, we see:
> 
> 
> $$\ce{Ba^{2+} + 2Cl- + 2Na^{+} + SO4^{2-} -> BaSO4 (s) + 2Na^{+} + 2Cl^{-}}$$
> 
> 
> $\ce{Na^{+}, Cl^{-}}$ remains completely dissociated in the aqueous solution as they're salts of strong acid and strong base (i.e., they're strong electrolytes). Hence, $\ce{Na^{+}, Cl^{-}}$ are spectator ions as they don't participate in ionic reaction. Let's cancel them out, and we get the net ionic reaction:
> 
> 
> $$\ce{Ba^{2+} + SO4^{2-} -> BaSO4 (s)}$$
> 
> 
> 


Now, let's see the reaction given in the question. In this, $\ce{BaSO4}$ precipitates out, so $\ce{Ba^{2+}, SO4^{2-}}$ can't be spectator ions. Also, H2O has a very low dissociation contant ($\ce{K\_W = 10^{-14}}$, at 25°C), so it can't dissociate into ions. Hence, $\ce{H+, OH-}$ are too not the spectator ion. Therefore, this reaction has no spectator ion and hence **${H\_2O}$ will be included in the ionic reaction**, which is given as:


$$\ce{Ba^{2+} + 2OH- + 2H+ + SO4^{2-} -> BaSO4 (s) + 2H2O(l)}$$




---


Another example:


**[1]:** [Complete ionic and net ionic equations, Khan Academy](https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/types-of-chemical-reactions/a/complete-ionic-and-net-ionic-equations)


Note: As mentioned by andselisk in comments, $\ce{BaSO4 (s)}$ is the recommended way to write the precipitate, as per IUPAC. Old school symbol for precipitate was "↓".


 3

Since you don't seem to understand the concepts let's break this down more. Rahul's answer is correct, but let's go into more detail. Consider dissolving barium chloride, equation 1, and sodium sulfate,equation 2, in separate solutions. (*Note that it is important to show the charge on ions!*)


$$\ce{BaCl2(s) ->[aq] Ba^{2+}(aq) + 2Cl-(aq)}\tag{1}$$
$$\ce{Na2SO4(s) ->[aq] 2Na^{+}(aq) + SO4^{2-}(aq)}\tag{2}$$


Now if the two ionic solutions are mixed the reaction is:
$$\ce{Ba^{2+}(aq) + 2Cl-(aq) + 2Na^{+}(aq) + SO4^{2-}(aq) <=> BaSO4(s) + 2Na^{+}(aq) + 2Cl-(aq)}\tag{3}$$


* (s) means that the reactant is a solid, or that the product is a precipitate.
* (aq) means that the molecule or ion is dissolved in water. Often this it is assumed that the context is an aqueous solution and the designation is just left off.


Since the sodium ions and the chloride ions appear on both sides of the equation they are often canceled out as spectator ions. So for a reaction you might see an equation like equation (4). Typically one would assume that for equation 4 that the $\ce{Ba^{2+}}$ and $\ce{SO4^{2-}}$ ions came from some unspecified soluble salts, and that the reaction was happening in water. 


$$\ce{Ba^{2+} + SO4^{2-} <=> BaSO4(s)}\tag{4}$$


Another way to specify reaction conditions would be to put such specifications over the reaction arrows. Such as using $\ce{ ->[CH3OH]}$ to indicate that the reaction is happening in methanol. 


 2