**Predicting Whether a Salt Solution is Acidic, Basic, or Neutral:**

**Example:**

Salt: NaCl

Parents: NaOH/HCl

pH: neutral (a strong acid and a strong base)

Salt: NaF

Parents: NaOH/HF

pH: basic (strong base, weak acid)

Salt: NH

_{4}Br

Parents: NH

_{3}/HBr

pH: acidic (strong acid, weak base)

Salt: NH

_{4}NO

_{2}

Parents: NH

_{3}/HNO

_{2}

pH: must find K

_{a}, K

_{b}and compare

But what do we need to find the K

_{a}and K

_{b}for? When dealing with a salt, follow solubility rules. All salts of ammonium are soluble, so the salt will break apart in water:

NH

_{4}NO

_{2}(aq) → NH

_{4}

^{+}(aq) + NO

_{2}

^{-}(aq)

These ions will go on to react with water, where one will produce OH

^{-}(giving a basic property) and one will produce H

_{3}O

^{+}(to produce an acidic quality). Whether the solution itself is acidic or basic depends on whether more OH

^{-}is produced or more H

_{3}O

^{+}is produced. In other words, we can find whether the solution is acidic or basic by seeing which is larger, the K

_{a}or the K

_{b}:

NH

_{4}

^{+}(aq) + H

_{2}O (l) ⇌ H

_{3}O

^{+}(aq) + NH

_{3}(aq)

NO

_{2}

^{-}(aq) + H

_{2}0 (l) ⇌ OH

^{–}(aq) + HNO

_{2}(aq)

Because the reaction with NH

_{4}

^{+}produces H

_{3}O

^{+}, we know we need to find the K

_{a}for this reaction. The table doesn't list the K

_{a}for NH

_{4}

^{+}. How do we find it?

The product of K

_{a}and K

_{b}for conjugate acid-base pairs equals K

_{w}.

K

_{a}x K

_{b}= K

_{w}

So to find the Ka for NH

_{4}

^{+}, we find the K

_{b}of the conjugate base, and plug these numbers into the above equation to solve for K

_{a}.

K

_{aNH4+}= K

_{w}/ K

_{bNH3}= (1 x 10

^{-14}) / (1.8 x 10

^{-5}) = 5.56 x 10

^{-6}

Because the reaction with NO

_{2}

^{-}produces OH

^{-}, we know we need to find the K

_{b}for this reaction. The table doesn't list the K

_{b}for NO

_{2}

^{-}. How do we find it?

Again, find the K

_{a}for the conjugate acid, and plug these numbers into the above equation and solve for K

_{b}.

K

_{bNO2-}= K

_{w}/ K

_{aHNO2}= (1 x 10

^{-14}) / (4.5 x 10

^{-4}) = 2.22 x 10

^{-11}

Here, K

_{a}is greater than K

_{b}, so the solution will be acidic.

**Example:**

What is the pH for 0.250 M KClO?

KClO is a salt. According to solubility rules, it is soluble:

KClO (aq) → K

^{+}(aq) + ClO

^{-}(aq)

These ions will go on to react with water, where one will produce OH

^{-}(giving a basic property) and one will produce H

_{3}O

^{+}(to produce an acidic quality).

K

^{+}(aq) +H

_{2}O (l) ⇌ KOH (aq) + H

_{3}O

^{+}(aq)

ClO

^{-}(aq) + H

_{2}O (l) ⇌ HClO (aq) OH

^{-}(aq)

But there's a catch. KOH is a strong base, and we know that strong bases dissociate (ionize) 100% in water. So really,

K

^{+}(aq) + H

_{2}O (l) → No Reaction

We also know that we had 0.250 M KClO to begin with. The KClO dissociated into K

^{+}and ClO

^{-}where everything is 1:1, so it dissociates to form 0.250 M of K

^{+}(that produces no reaction) and 0.250 M ClO

^{-}, which will go on to react with water. This is a little more obvious when we build a table:

Now that we know we have 0.250M ClO

^{-}, we can build our table for the reaction with water:

ClO

^{-}(aq) + H

_{2}O (l) ⇌ HClO (aq) OH

^{-}(aq)

Hint: we know we need to find K

_{b}because OH

^{-}is a product

K

_{b}= (x

^{2}) / (0.250 – x) → approximation method holds

How do we find the K

_{b}for ClO

^{-}?

K

_{bClO-}= K

_{w}/ K

_{aHClO}= (1 x 10

^{-14}) / (3.5 x 10

^{-8}) = 2.89 x 10

^{-7}

So: 2.89 x 10

^{-7}= (x

^{2}) / 0.250

x = 2.69 x 10

^{-4}= [OH

^{-}]

pOH = -log(2.69 x 10

^{-4}) = 3.57

pH = 14 – 3.57 = 10.4

**Example:**

What is the pH for 0.135M NH

_{4}Cl?

NH

_{4}Cl is a salt, which ions will it form when it dissociates?

NH

_{4}

^{+}and Cl

^{-}

NH

_{4}Cl → NH

_{4}

^{+}+ Cl

^{-}

These ions with go on to react with water. We know Cl

^{-}will react to form HCl, which is a strong acid, and as such it will dissociate 100% in water, so overall, there is no reaction. So what happens to the NH

_{4}

^{+}? It is the conjugate acid of a weak base, so:

NH

_{4}

^{+}(aq) + H

_{2}O (l) ⇌ NH

_{3}(aq) + H

_{3}O

^{+}(aq)

Again, it is easy to see that since we start out with 0.135 M of NH

_{4}Cl, we will get 0.135 M of each of the ions (1:1), so when we build a table for the reaction of the ammonium ion with water, we know to start with 0.135 M.

Hint: we know we need to find K

_{a}because H

_{3}O

^{+}is a product

K

_{a}= (x

^{2}) / (0.135 – x) → approximation holds

How do we find the K

_{a}for NH

_{4}

^{+}?

K

_{aNH4+}= K

_{w}/ K

_{bNH3}= (1 x 10

^{-14}) / (1.8 x 10

^{-5}) = 5.56 x 10

^{-10}

So: 5.56 x 10

^{-10}= (x

^{2}) / (0.135)

x = 8.66 x 10

^{-6}= [H

_{3}O

^{+}]

pH = -log(8.66 x 10

^{-6}) = 5.06

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