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Question: How would you balance \[{H_3}P{O_4}(aq) + Ca{(OH)_2}(s) \to C{a_3}{(P{O_4})_2}(s) + {H_2}O(l)\]...

How would you balance
H3PO4(aq)+Ca(OH)2(s)→Ca3(PO4)2(s)+H2O(l){H_3}P{O_4}(aq) + Ca{(OH)_2}(s) \to C{a_3}{(P{O_4})_2}(s) + {H_2}O(l)

Explanation

Solution

Hint In order to balance the given chemical equation, we must first know what a balanced equation is. We should also have an idea about how to balance the chemical equation. When the number of atoms present in both the reaction and product side in a given chemical equation is equal. This is called the balanced chemical equation.

Complete step by step solution:
First let us understand what a chemical equation is. When a chemical reaction is represented in the chemical symbol, then it is called a chemical equation. A chemical equation will be having a reactant, product and an arrow to represent the direction. When the number of atoms present in both the reaction and product side in a given chemical equation is equal. This is called the balanced chemical equation. Now let us move onto the given question.
Ca(OH)2Ca{(OH)_2} is a very strong base which will react with H3PO4{H_3}P{O_4} which is a very weak acid. They both will undergo a neutralisation reaction to give Ca3(PO4)2C{a_3}{(P{O_4})_2} an insoluble salt and water.
H3PO4(aq)+Ca(OH)2(s)→Ca3(PO4)2(s)+H2O(l){H_3}P{O_4}(aq) + Ca{(OH)_2}(s) \to C{a_3}{(P{O_4})_2}(s) + {H_2}O(l)
The above given reaction is an unbalanced equation, we have to balance it.
We can balance the above equation by using the ions. First let us take phosphoric acid.
H3PO4(aq)→3H+(aq)+PO43−(aq){H_3}P{O_4}(aq) \to 3{H^ + }(aq) + PO_4^{3 - }(aq)
Now we have to take calcium hydroxide.
Ca(OH)2(aq)→Ca2+(aq)+2OH−(aq)Ca{(OH)_2}(aq) \to C{a^{2 + }}(aq) + 2O{H^ - }(aq)
Now we have to take the calcium phosphate
Ca3(PO4)2(aq)→3Ca+(aq)+2PO43−(aq)C{a_3}{(P{O_4})_2}(aq) \to 3C{a^ + }(aq) + 2PO_4^{3 - }(aq)
Write the whole unbalanced equation as
3H+(aq)+PO43−(aq)+Ca2+(aq)+2OH−(aq)→3Ca2+(aq)+2PO43−(aq)+H2O(l)3{H^ + }(aq) + PO_4^{3 - }(aq) + C{a^{2 + }}(aq) + 2O{H^ - }(aq) \to 3C{a^{2 + }}(aq) + 2PO_4^{3 - }(aq) + {H_2}O(l)
Now all we have to do is to balance out the ions on both sides.
So, in order to get three calcium ions on the reactant side, we have to multiply the calcium hydroxide by 3. To get two phosphate ions on the reactant side, we have to multiply the phosphoric acid by 2.
This will give you
6H+(aq)+2PO43−(aq)+3Ca2+(aq)+6OH−(aq)→3Ca2+(aq)+2PO43−(aq)+H2O(l)6{H^ + }(aq) + 2PO_4^{3 - }(aq) + 3C{a^{2 + }}(aq) + 6O{H^ - }(aq) \to 3C{a^{2 + }}(aq) + 2PO_4^{3 - }(aq) + {H_2}O(l)
Finally, to balance the hydrogen and oxygen on both sides, we have to multiply the water molecule by 6.
6H+(aq)+2PO43−(aq)+3Ca2+(aq)+6OH−(aq)→3Ca2+(aq)+2PO43−(aq)+6H2O(l)6{H^ + }(aq) + 2PO_4^{3 - }(aq) + 3C{a^{2 + }}(aq) + 6O{H^ - }(aq) \to 3C{a^{2 + }}(aq) + 2PO_4^{3 - }(aq) + 6{H_2}O(l)
Finally, we have to write the balanced the equation:
2H3PO4(aq)+3Ca(OH)2(s)→Ca3(PO4)2(s)+6H2O(l)2{H_3}P{O_4}(aq) + 3Ca{(OH)_2}(s) \to C{a_3}{(P{O_4})_2}(s) + 6{H_2}O(l)

Note: When we are writing a chemical equation, we should never forget to balance the equation because when the chemical equation is unbalanced, then the equation is incorrect or incomplete. Therefore, always remember to balance the chemical equation.