# Which of the given salt will have maximum pH?

## ${\text{KNO}}_{3}$ $\text{NaCN}$ ${\text{HCOONH}}_{4}$ $\text{NaOH}$

Apr 8, 2018

2.

#### Explanation:

As people pointed out, it is good to first know the definition of a salt:
An ionic compound formed by the neutralization of an acid and a base
- That eliminates 4. As NaOH is not achieved through neutralization. (But it is often used as a reagent in a neutralization reaction)

Compound 1,however, is formed in the neutralization reaction between Potassium hydroxide and Nitric acid:
$K O H \left(a q\right) + H N {O}_{3} \left(a q\right) \to K N {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$
But it is not particularly exciting to dissolve- it simply dissociates into ions, ${K}^{+}$ and $N {O}_{3}^{-}$, neither of which affects the $p H$.

Compound 3 on the other hand, is formed by neutralizing methanoic acid with ammonia, and can dissociate into the ammonium ion $\left(N {H}_{4}^{+}\right)$ and methanoate ion $\left(H C O {O}^{-}\right)$. Now, the ammonium ion is a weak acid and will react with Water to form Oxonium ions,
$N {H}_{4}^{+} \left(a q\right) + {H}_{2} O \left(l\right) \to N {H}_{3} \left(a q\right) + {H}_{3} {O}^{+} \left(a q\right)$
which will decrease the $p H$!- This was the opposite of what we were looking for.

Finally, we have compound 2.
This is formed from neutralizing Hydrogen cyanide, $H C N$, which is a weak acid with Sodium hydroxide $N a O H$, a strong base.

The cyanide ion, $C {N}^{-}$, is, therefore, the conjugate base of $H C N$ in the (HCN) / (CN^-) acid/base pair.
So if we put $N a C N$ in water, it will dissociate to form $N {a}^{+}$ and $C {N}^{-}$, and $C {N}^{-}$ will act as a Bronstedt-Lowry base and accept a proton from water:

$N a C N \left(s\right) + {H}_{2} O \left(l\right) \to N {a}^{+} \left(a q\right) + O {H}^{-} \left(a q\right) + H C N \left(a q\right)$
Which forms hydroxide ions! These will increase the $p H$ of the solution above 7, so this salt will give the maximum $p H$!