pH - GRE

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Question

HCN dissociates based on the following reaction.

$HCN + H_{2}O \rightarrow CN^{-}+ H_{3}O^{+}$

The Ka for hydrogen cyanide is $\small 6.2*10^{-10}$ .

of is added to of water. What is the pH of the resulting solution?

Answer

Since HCN is a weak acid, we must use the equilibrium equation.

$K_{a}= \frac{[CN^{-}][H^{+}]}{[HCN]}$

Because the HCN dissociates in solution, we expect the concentrations of protons and cyanide ions to increase, while the concentration of HCN will decrease. After determining the molarity of the solution, we can set up the equation below, using X as the amount of moles that dissociate.

$\small K_{a}= \frac{[X][X]}{[[\frac{0.4mol}{2L}]-X]} = 6.210^{-10}$

Because X is small, we can neglect its impact in the denominator.

$\small K_{a}= \frac{[X][X]}{[\frac{0.4mol}{2L}]} = \frac{[X]^2}{[0.2]}=6.210^{-10}$

$\small X = 1.110^{-5}$

Since X is the concentration of protons in the solution, we can calculate the pH by using the equation $pH = -log[H^{+}]$ .

$pH=-log[1.110^{-5}]=4.95$

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Question

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

Answer

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.

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Question

What is the pH of a $5.0*10^{-4} M$ solution of $Ca(OH)_{2}$ ?

Answer

First we need to calculate the of the $510^{-4} M$ $Ca(OH)_{2}$ solution. For every mole of $Ca(OH)_{2}$ , there is double the number of moles of hydroxide ions:

$[OH^{-}]=2(5 10^{-4} M)=0.001M$

The pOH can be calculated using the below equation:

$pOH=-log(OH^{-})=-log(0.001)=3$

The equation with the relationship between pH and pOH is below:

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:

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.010^{-14}$

Rearrange this equation:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.010^{-14}}{0.001}=110^{-11}$

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

$pH=-log(H_{3}O^{+})=-log(1.0*10^{-11})=11$

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Question

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

Answer

We need to calculate the pH of a $210^{-4} M$ solution. There is one mole of in every mole of , therefore:

$[OH^{-}]=210^{-4}\ M$

$pOH=-log(OH^{-})=-log(210^{-4})=3.7$

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:

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.010^{-14}$

Rearrange this equation:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.010^{-14}}{210^{-4}}=510^{-11}$

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

$pH=-log(H_{3}O^{+})=-log(5.010^{-11})=10.3$

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Question

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

Answer

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:

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

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Question

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

Answer

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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Question

What is the pH of a $0.10 \textup{ M}$ solution of ?

Answer

Below is the equilibria of in an aqueous solution:

$NaOH(s)\rightleftharpoons\ Na^{+}(aq)+OH^{-}(aq)$

is a strong base so it completely ionizes in solution. It has a high solubility product constant.

There is 100% dissociation of , therefore the resulting $OH^{-}$ in solution equals to the concentration of original .

$[OH^{-}]=0.10M$

$pOH=-log[OH^{-}]=-log[0.10]=1$

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Question

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 ?

Answer

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 .

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

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

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

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Question

Find the pH of $H_{2}O$ if its $K_{w}$ is $5.476\times10^{-14}$ at $50^{o}C$ .

Answer

$K_{w}$ is a symbol for the ionization constant for water. This value is temperature dependent. At $50^{o}C$ , the $K_{w}$ for water is $5.476\times10^{-14}$ .

Water dissociates or ionizes according to the chemical equation below:

$H_{2}O(l)\rightleftharpoons\ H^{+}(aq)+OH^{-}(aq)$

Therefore, the ionization of water results in:

$[H^{+}]=[OH^{-}]$

$K_{w}=[H_{3}O^{+}][OH^{-}]$

Let us set $x=[H^{+}]=[OH^{-}]$

$K_{w}=x^{2}$

$\sqrt{K_{w}}=x$

$\sqrt{5.476\times10^{-14}}=x$

$x=2.340\times10^{-7}\ M$

$pH=-log[H^{+}]=-log(2.340\times10^{-7}\)=6.63$

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Question

What is the $H^{+}$ concentration of a solution with pH ?

Answer

The pH of a solution is related to the number of hydrogen ions in solutions. The equation to determine the pH of a solution is below:

$pH=-log[H^{+}]$

The above equation can be converted into the following form in order to determine the hydrogen ion concentration:

$[H^{+}]=10^{-pH}$

Therefore, the hydrogen ion concentration of a solution of pH 7 is:

$[H^{+}]=10^{-7}=1.0\times10^{-7}M$

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Question

What is the pH of a solution with a pOH of ?

Answer

The equation that expresses the relationship between pH and pOH is:

Plugging the pOH given into the equation above gives:

Rearranging this expression gives:

Therefore, the pH of a solution with a pOH of is:

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Question

What is the pH of a solution in which $H^{+}$ concentration is $1.2\times10^{-8}M$ ?

Answer

The pH of a solution is related to the number of hydrogen ions in solutions. The equation to determine the pH of a solution is below:

$pH=-log[H^{+}]$

Plugging the concentration given into the equation above gives:

$pH=-log[1.2\times 10^{-8}M]$

Therefore, the pH of the solution is:

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Question

Considering the $K_{a}$ for HCN is $4.9\times10^{-10}$ , what is the $K_{b}$ for $CN^{-}$ ?

Answer

The equilibrium governing the dissolution of HCN in water is:

$HCN_{(aq)}+H_{2}O_{(l)}\rightleftharpoons\ CN^{-}_{(aq)}+H_{3}O^{+}_{(aq)}$

is the conjugate acid of $CN^{-}$ . In other words, $CN^{-}$ is the conjugate base of .

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

By rearranging the equation we get:

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{4.9\times 10^{-10}}=2.0\times10^{-5}$

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Question

What is the pH of a $7.1\times10^{-4}\ M$ solution of ?

Answer

We need to calculate the pH of a $7.1\times10^{-4}\ M$ solution. There is one mole of $OH^{-}$ in every mole

of , therefore:

$[OH^{-}]= 7.1\times10^{-4}\ M$

$pOH=-log(OH^{-})=-log(7.1\times10^{-4}\)=3.1$

The equation with the relationship between pH and pOH is below:

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:

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.0\times10^{-14}$

Rearranging this equation gives:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0\times 10^{-14}}{7.1\times 10^{-4}}=1.4\times10^{-11}$

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

$pH=-log[H_{3}O^{+}]=-log(1.4\times10^{-11}\)=10.9$

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Question

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

Answer

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:

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

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Question

What is the pH of an aqueous solution with a $5.1\times10^{-6}\ M$ hydroxide ion concentration?

Answer

We need to calculate the pH of a solution with a $5.1\times10^{-6}\ M$ hydroxide ion concentration.

$[OH^{-}]= 5.1\times10^{-6}\ M$

$pOH=-log(OH^{-})=-log(5.1\times10^{-6}\)=5.3$

The equation with the relationship between pH and pOH is below:

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:

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.0\times10^{-14}$

Rearranging this equation gives:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0\times 10^{-14}}{5.1\times10^{-6}\ }=1.96\times10^{-9}$

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

$pH=-log[H_{3}O^{+}]=-log(1.96\times10^{-9}\)=8.7$

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Question

Determine the pH of a $0.10\ M$ solution.

Answer

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

$HCl(aq)\rightarrow\ H^{+}(aq)+Cl^{-}(aq)$

This means that the $[HCl]=[H^{+}]\ in\ solution$

Below is the equation to calculate pH:

$pH=-log[H^{+}]$

Therefore,

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Question

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

Answer

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:

$KOH(aq)\rightarrow\ K^{+}(aq)+OH}^{-}(aq)$

This means that the $[KOH]=[OH^{-}]\ in\ solution$

NaOH solution. There is one mole of OH- in every mole

of NaOH, therefore:

$[OH^{-}]= 0.12\ M$

$pOH=-log(OH^{-})=-log(0.12)=0.92$

The equation with the relationship between pH and pOH is below:

We can calculate the pH by rearranging this equation:

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Question

Find the pH of $H_{2}O$ if its $K_{w}$ is $1.0\times10^{-14}$ at $25^{o}C$ .

Answer

$K_{w}$ is a symbol for the ionization constant for water. This value is temperature dependent. At $25^{o}C$ , the $K_{w}$ for water is $8.8\times10^{-14}$ .

Water dissociates or ionizes according to the chemical equation below:

$H_{2}O(l)\rightleftharpoons\ H^{+}(aq)+OH^{-}(aq)$

Therefore, the ionization of water results in:

$[H^{+}]=[OH^{-}]$

$K_{w}=[H_{3}O^{+}][OH^{-}]$

Let us set $x=[H^{+}]=[OH^{-}]$

$K_{w}=x^{2}$

$\sqrt{K_{w}}=x$

$\sqrt{1.0\times 10^{-14}}=x$

$x=1.0\times10^{-7}\ M$

$pH=-log[H^{+}]=-log(1.0\times10^{-7}\)=7$

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Question

Determine the pH of a $0.02\ M$ $HNO_{3}$ solution.

Answer

$HNO_{3}$ is a strong acid and therefore completely ionizes in solution. Therefore, all the $H^{+}$ ions dissociate in the $HNO_{3}$ molecules in solutions according to the below chemical equation:

$HNO_{3}(aq)\rightarrow\ H^{+}(aq)+NO_{3}^{-}(aq)$

This means that the $[HNO_{3}]=[H^{+}]\ in\ solution$

Below is the equation to calculate pH:

$pH=-log[H^{+}]$

Therefore,

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