GRE Subject Test: Chemistry - pH

Considering the $K_{a}$ for acetic acid $(CH_{3}COOH)$ is $1.8\times10^{-5}$ , what is the $K_{b}$ for acetate $(CH_{3}COO^{-})$ ?

$5.6\times10^{-10}$

$3.5\times10^{-4}$

$1.1\times10^{-8}$

$3.0\times10^{-3}$

Explanation

The equilibrium governing the dissolution of $CH_{3}COOH$ in water is:

$CH_{3}COOH_{(aq)}+H_{2}O_{(l)}\rightleftharpoons\ CH_{3}COO^{-}_{(aq)}+H_{3}O^{+}_{(aq)}$

$CH_{3}COOH$ is the conjugate acid of $CH_{3}COO^{-}$ . In other words, $CH_{3}COO^{-}$ is the conjugate base of $CH_{3}COOH$ .

Using the relationship, $K_{w}=K_{a}K_{b}$ , we can calculate the Kb.

By rearranging the equation we get:

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.6\times10^{-10}$

Considering the of (hydrofluoric acid) is $6.8\times10^{-4}$ , what is the of the base ?

Explanation

The relationship between and is:

$K_{a}*\ K_{b}=K_{w}$

$K_{w}=1.0\times10^{-14}$

Rearranging this equation gives:

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{6.8\times 10^{-4}}=1.5\times10^{-11}$

In order to calculate the , we must use this relationship:

$pK_{b}=-log(K_{b})=-log(1.5\times 10^{-11})=10.8$

What is the molarity of a solution that has a pH of ?

$1 * 10^{-5} M$

$5.5 * 10^{-1} M$

Explanation

The pH of the solution is , therefore the concentration would be $1 * 10^{-5} M$ . This solution is based on the equation, $[H^{+}] = 10^{-pH}$ and because hydrochloric is a strong acid, it can be assumed to completely dissociate in solution.

What is the pH of a $2.0*10^{-4} M$ solution of ?

Explanation

We need to calculate the pH of a $2*10^{-4} M$ solution. There is one mole of in every mole of , therefore:

$[OH^{-}]=2*10^{-4}\ M$

$pOH=-log(OH^{-})=-log(2*10^{-4})=3.7$

The equation with the relationship between pH and pOH is:

We can calculate the pH by rearranging this equation:

Another way of solving this problem is shown below. The equation with the relationship between $H_{3}O^{+}$ and $OH^{-}$ concentration is:

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.0*10^{-14}$

Rearrange this equation:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0*10^{-14}}{2*10^{-4}}=5*10^{-11}$

We can calculate the pH of this solution using the equation below:

$pH=-log(H_{3}O^{+})=-log(5.0*10^{-11})=10.3$

What is the pH of a $0.05 \textup{ M}$ solution of ?

Explanation

Below is the equilibria of in an aqueous solution:

$HCl(l) +H_{2}O(l)\rightleftharpoons\ Cl^{-}(aq)+H_{3}O^{+}(aq)$

is a strong acid so it completely ionizes in solution. It has a high $K_{eq}$ of $1.6\times10^{6}$ .

There is 100% dissociation of , therefore the resulting $H_{3}O^+$ in solution equals to the concentration of original .

$[H_{3}O^{+}]=0.05M$

$pH=-log[H_{3}O^{+}]=-log[0.05]=1.3$

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If $3.11\times10^{-4}\ moles$ of is reacted with $5.71\times10^{-4}\ moles$ of , what is the pH if the total volume of the solution is 1 liter ?

Explanation

Based on the chemical equation, and react in a 1:1 mole ratio. Therefore, in order to find the pH of this solution you must first determine the difference in moles of the two reactants and the limiting reactant. The reactant in excess will determine the pH of the solution. $5.71\times 10^{-4}\ moles\ HCl-3.11\times 10^{-4}\ moles\ NaOH=2.59\times 10^{-4}\ moles\ HCl$

There is in excess and the pH of the solution will be based on the moles of the excess .

$moles\ of\ H^{+}=2.59\times 10^{-4}}\ moles$

$Molarity\ [H^+]=\frac{moles\ of\ solute}{Liters\ of\ solution}=\frac{2.6\ moles}{1L}=2.6M$

$pH=-log[H^{+}]=-log[2.59\times 10^{-4}]=3.6$

Calculate the Ka for $HNO_{2}$ .

$5.0\times10^{-4}$

$2.5\times10^{-4}$

$5.6\times10^{-12}$

Explanation

The pka of $HNO_{2}$ is . The pka is simply calculated based on the following equation:

Therefore, in order to determine the ka, simply use f the following equation:

$ka=10^{-pka}$

$ka=10^{-3.3}$

$ka=5.0\times10^{-4}$

Determine the pH of a $0.12\ M$ solution of .

Explanation

is a strong base and therefore completely ionizes in solution. Therefore, all the $OH^{-}$ ions dissociate in the molecules in solutions according to the below chemical equation: