Question

# When borax is dissolved in water:A.$B{\left( {OH} \right)_3}$ is formed onlyB.${\left[ {B{{\left( {OH} \right)}_4}} \right]^ - }$ formed onlyC.Both $B{\left( {OH} \right)_3}$ and ${\left[ {B{{\left( {OH} \right)}_4}} \right]^ - }$ are formedD.${\left[ {{B_2}{O_3}{{\left( {OH} \right)}_4}} \right]^ - }$ is formed only

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Hint: We have to know that borax is nothing but sodium borate, sodium tetraborate, (or) disodium tetraborate. It is a compound that has a chemical formula $N{a_2}{H_4}{B_4}{O_9} \cdot n{H_2}O$. It is a crystalline solid that appears colourless, which is soluble in water. Borax is produced from other boron compounds and recrystallization is the process for refining naturally occurring borax.

We have to know that the borax is used to represent the amount of related minerals (or) chemical compounds, which varies in their crystal water content. The formula of anhydrous sodium tetraborate is $N{a_2}{B_4}{O_7}$. The formula of sodium tetraborate pentahydrate is $N{a_2}{B_4}{O_7} \cdot 5{H_2}O$. The formula of sodium tetraborate is $N{a_2}{B_4}{O_7} \cdot 10{H_2}O$ (or) similarly the formula of sodium octahydrate is $N{a_2}{B_4}{O_7} \cdot 8{H_2}O$.
We have to that borax contains ${\left[ {{B_4}{O_5}{{\left( {OH} \right)}_4}} \right]^{2 - }}$ ion. In this structure of borax, the number of four-coordinate boron centers is two and the number of three-coordinate boron centers is two. When borax is dissolved in water, $B{\left( {OH} \right)_3}$ and ${\left[ {B{{\left( {OH} \right)}_4}} \right]^ - }$. We can write the chemical equation of boron dissolved in water as,
${\left[ {{B_4}{O_5}{{\left( {OH} \right)}_4}} \right]^{2 - }} + 5{H_2}O \to 2B{\left( {OH} \right)_3} + 2{\left[ {B{{\left( {OH} \right)}_4}} \right]^ - }$
The presence of $B{\left( {OH} \right)_3}$ and ${\left[ {B{{\left( {OH} \right)}_4}} \right]^ - }$ ensures that borax comprises of two triangular units of $B{O_3}$ and two tetrahedral units of $B{O_4}$ units.
When we use borax for titrating against acids, only ${\left[ {B{{\left( {OH} \right)}_4}} \right]^ - }$ reacts with hydrochloric acid, $B{\left( {OH} \right)_3}$ is formed. The chemical equation is written as,
$2{\left[ {B{{\left( {OH} \right)}_4}} \right]^ - } + 2{H_3}{O^ + } \to 2B{\left( {OH} \right)_3} + 4{H_2}O$
The product $B{\left( {OH} \right)_3}$ is formed when ${\left[ {B{{\left( {OH} \right)}_4}} \right]^ - }$ is reacted with hydrochloric acid. Option (A), (B), and (D) are incorrect when borax is dissolved in water.