# Question #42e61

May 31, 2015

$2 N a O H + {H}_{2} S {O}_{4} \to N {a}_{2} S {O}_{4} + 2 H O H$

This is an example of a neutralization (double decomposition) reaction between a base and an acid which usually produces a salt and water.

$$ It is graphically represented as:


$\text{B-OH" + "H-A" -> "B-A" + "H"_2"O}$

where $B$ is the metal in the base (with the $O H$) and $A$ is the nonmetal /radical in the acid (with H as the positive ion) resulting to a binary/ternary salt, $B - A$.

It is called binary because only two elements make it up, a metal as the positive ion and the nonmetal as the negative ion.

Ternary salts are made up of a metal with a radical (a group of elements exhibiting an electrical charge) , e.g., $S {O}_{4}^{- 2}$

Water is formed from the positive ion of the acid and the negative $O {H}^{-}$ radical of the base.

$$ Therefore, to form the salt, just exchange the electrical charges of the metal and the radical:


$2 N {a}^{+} + S {O}_{4}^{2 -} \to N {a}_{2} S {O}_{4}$