# Assuming equal concentrations, which of these salt solutions should have the highest pH? (Choices in answer).

Aug 4, 2015

The answer is D. potassium nitrite

#### Explanation:

The options are:

A. $C a {\left(N {O}_{3}\right)}_{\textrm{2 \left(a q\right]}}$

B. $F e C {l}_{3 \left(a q\right)}$
C. $N {a}_{2} S {O}_{4 \left(a q\right)}$
D. $K N {O}_{2 \left(a q\right)}$
E. $N {H}_{4} {F}_{\left(a q\right)}$

Once again, you can use equilibrium reactions to determine which of those salts will produce a solution with an acidic pH.

The idea here is that when you dissolve a soluble salt in water, it dissociates into cations and anions.

These cations and anions can react with water molecules to affect the pH of the solution by increasing the concentration of hydronium ions or hydroxide ions.

The cations of strong bases and the anions of strong acids will be neutral because the equilibrium reaction that describes the dissociation of strong acids and strong bases lies so far to the right that you can assume complete dissociation.

This means that the cations of strong bases are very weak conjugate acids (weaker than water) and the anions of strong acids are very weak bases (weaker than water), and thus cannot react with water to reform the strong acid or strong base.

Calcium nitrate and sodium sulfate will form neutral solutions, since $C {a}^{2 +}$ and $N {a}^{+}$ are the cations of strong bases ($C a {\left(O H\right)}_{2}$ and $N a O H$, respectively) and NO_3""^(-) and $S {O}_{4}^{2 -}$ are the anions of strong acids ($H N {O}_{3}$ and ${H}_{2} S {O}_{4}$, respectively).

Iron (III) chloride will form an acidic solution because the iron (III) cation will react with water to form hydronium ions. That happens because cations that are smaller in size and have high charges will act as weak acids.

$F e {\left({H}_{2} O\right)}_{6 \left(a q\right)}^{3 +} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + F e {\left({H}_{2} O\right)}_{5} O {H}_{\textrm{\left(a q\right]}}^{2 +}$

Ammonium fluoride is a little more complicated because both the cation and the anion will hydrolyze. In such cases, you need to compare the strength of the cation as an acid with the strength of the anion as a base.

Since $N {H}_{4}^{+}$ has $p {K}_{a} = 9.25$ and ${F}^{-}$ has $p {K}_{b} = 10.8$, the solution will be acidic because the ammonium ion is a stronger acid than the fluoride anion is a base, i.e. it ionizes more readily.

Finally, potassium nitrite will produce a basic solution because the nitrite anion, NO_2""^(-), will react with water to form nitrous acid, $H N {O}_{2}$, a weak acid, and hydroxide ions.

$N {O}_{2 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s H N {O}_{2 \left(a q\right)} + O {H}_{\left(a q\right)}^{-}$

Thus, the solution with the highest pH will be the potassium nitrite solution.