I need to find the ionic equation of:
3Ba(OH)2(aq)+2H3PO4(aq)->Ba3(PO4)2+6H20(l)
Well, let me see if I can find some humor in this equation for you.
Why did the barium hydroxide and phosphoric acid go on a date?
Because they wanted to create some "Ba-PO" chemistry!
Okay, now let's break it down into an ionic equation:
3Ba(OH)2(aq) + 2H3PO4(aq) -> Ba3(PO4)2(s) + 6H2O(l)
In the reaction, the barium hydroxide (Ba(OH)2) molecules split into three Ba2+ ions and six OH- ions, while the phosphoric acid (H3PO4) molecules split into two H+ ions and one phosphate ion (PO43-).
So, the ionic equation becomes:
3Ba2+(aq) + 6OH-(aq) + 2H+(aq) + PO43-(aq) -> Ba3(PO4)2(s) + 6H2O(l)
Hope that brings a little joy to your equation-balancing journey!
To find the ionic equation, we need to start by writing out the formula for each compound and then breaking them down into their constituent ions.
The given equation is:
3Ba(OH)2(aq) + 2H3PO4(aq) -> Ba3(PO4)2 + 6H2O(l)
First, let's write out the formulas for each compound:
1. Barium hydroxide: Ba(OH)2 (aq)
2. Phosphoric acid: H3PO4 (aq)
3. Barium phosphate: Ba3(PO4)2
4. Water: H2O (l)
Next, let's break down each compound into its ions:
1. Ba(OH)2 (aq) --> Ba^2+ (aq) + 2OH^- (aq)
2. H3PO4 (aq) --> 3H^+ (aq) + PO4^3- (aq)
Now, we can rewrite the balanced equation in terms of the ions:
3Ba^2+ (aq) + 6OH^- (aq) + 2 * 3H^+ (aq) + 2PO4^3- (aq) -> Ba3(PO4)2 + 6H2O (l)
Simplifying the equation, we get the ionic equation:
3Ba^2+ (aq) + 6OH^- (aq) + 6H^+ (aq) + 2PO4^3- (aq) -> Ba3(PO4)2 + 6H2O (l)
So, the ionic equation for the given reaction is:
3Ba^2+ (aq) + 6OH^- (aq) + 6H^+ (aq) + 2PO4^3- (aq) -> Ba3(PO4)2 + 6H2O (l)
To find the ionic equation for the given chemical reaction, you need to first write the balanced chemical equation. Once you have the balanced equation, you can separate the soluble ionic compounds into their respective ions and remove the spectator ions from the equation to write the ionic equation.
Let's start by writing the balanced chemical equation for the given reaction:
3Ba(OH)2(aq) + 2H3PO4(aq) -> Ba3(PO4)2(s) + 6H2O(l)
Now, let's break down the soluble compounds into their individual ions. In this reaction, Ba(OH)2 and H3PO4 are soluble compounds. The compounds will dissociate into their respective ions:
3Ba(OH)2(aq) breaks down into 6Ba2+(aq) + 6OH-(aq)
2H3PO4(aq) breaks down into 6H+(aq) + 2PO4^3-(aq)
Now, let's write the ionic equation by replacing the soluble compounds with their respective ions:
6Ba2+(aq) + 6OH-(aq) + 6H+(aq) + 2PO4^3-(aq) -> Ba3(PO4)2(s) + 6H2O(l)
Finally, you can simplify the equation by removing the spectator ions. In this case, Ba2+ and OH- are spectator ions present on both sides of the equation. Removing them leaves us with the ionic equation:
6H+(aq) + 2PO4^3-(aq) -> Ba3(PO4)2(s) + 6H2O(l)
So, the ionic equation for the given chemical reaction is 6H+(aq) + 2PO4^3-(aq) -> Ba3(PO4)2(s) + 6H2O(l).