Question

Suppose if I heat the products obtained on neutralising aqua regia with a mixture of NaOH and KOH in the ratio 1:3 (as it itself is a mixture of HCl and HNO₃ in the ratio 1:3) to any temperature more (not less) than 100 degree Celsius (or 212 degree Fahrenheit) then can I get the 2 salts (NaNO₃ & 3KCl) without 4H₂O as it boils at that temperature?

Answer 1

According to the theory at ordinary temperature: 

alkali + acid = salt + water. (reaction-exchange)

3HCl +HNO₃+NaOH+3KOH=NaNO₃ + 3KCl+ 4H ₂ O

The acid and alkali in the solution exchange ions and mutually neutralize each other.

The essence of the neutralization reaction is the interaction of protons with hydroxide ions, resulting in the formation of weakly dissociating molecules: H⁺+OH ^-⇔ H2O

Therefore, in this reaction, salts and water will be formed

But when acid and alkali are heated more (not less) than 100 degree Celsius , the water molecules are split off, it should form only the salt but not water (the salt in its anhydrous form):

aqua regia+mixture of NaOH and 3KOH=NaNO₃+3KCl