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

\(\displaystyle [OH^{-}]=2*10^{-4}\ M\)

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

The equation with the relationship between pH and pOH is:

\(\displaystyle pH+pOH=14\)

We can calculate the pH by rearranging this equation: 

\(\displaystyle pH= 14.00-3.7=10.3\)

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

\(\displaystyle K_{w}=[H_{3}O^{+}][OH^{-}]=1.0*10^{-14}\)

Rearrange this equation:

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

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

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

First we need to calculate the \(\displaystyle [OH^-]\) of the \(\displaystyle 5*10^{-4} M\) \(\displaystyle Ca(OH)_{2}\) solution. For every mole of \(\displaystyle Ca(OH)_{2}\) , there is double the number of moles of hydroxide ions:

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

The pOH can be calculated using the below equation:

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

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

\(\displaystyle pH+pOH=14\)

We can calculate the pH by rearranging this equation: 

\(\displaystyle pH= 14.00-3.00=11\)

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

\(\displaystyle K_{w}=[H_{3}O^{+}][OH^{-}]=1.0*10^{-14}\)

Rearrange this equation:

\(\displaystyle [H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0*10^{-14}}{0.001}=1*10^{-11}\)

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

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

The relationship between \(\displaystyle K_a\) and \(\displaystyle K_b\) is:

\(\displaystyle K_{a}*\ K_{b}=K_{w}\) 

\(\displaystyle K_{w}=1.0\times10^{-14}\)

Rearranging this equation gives:

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

In order to calculate the \(\displaystyle pK_b\), we must use this relationship:

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

Below is the equilibria of \(\displaystyle HCl\) in an aqueous solution:

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

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

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

\(\displaystyle [H_{3}O^{+}]=0.05M\)

\(\displaystyle pH=-log[H_{3}O^{+}]=-log[0.05]=1.3\)

Below is the equilibria of \(\displaystyle NaOH\) in an aqueous solution:

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

\(\displaystyle NaOH\) is a strong base so it completely ionizes in solution. It has a high solubility product constant.

There is 100% dissociation of \(\displaystyle NaOH\) , therefore the resulting \(\displaystyle OH^{-}\) in solution equals to the concentration of original \(\displaystyle NaOH\).

\(\displaystyle [OH^{-}]=0.10M\)

\(\displaystyle pOH=-log[OH^{-}]=-log[0.10]=1\)

\(\displaystyle pH+pOH=14\)

\(\displaystyle pH=14-pOH=14-1=13\)

Based on the chemical equation, \(\displaystyle HCl\) and \(\displaystyle NaOH\) 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.\(\displaystyle 5.71\times 10^{-4}\ moles\ HCl-3.11\times 10^{-4}\ moles\ NaOH=2.59\times 10^{-4}\ moles\ HCl\)

There \(\displaystyle HCl\) is in excess and the pH of the solution will be based on the moles of the excess \(\displaystyle HCl\).

\(\displaystyle moles\ of\ H^{+}=2.59\times 10^{-4}}\ moles\) 

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

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

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

\(\displaystyle [OH^{-}]= 5.1\times10^{-6}\ M\)

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

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

\(\displaystyle pH+pOH=14\)

We can calculate the pH by rearranging this equation:

\(\displaystyle pH=14.0-5.3=8.7\)

Another way of solving this problem is shown below. The equation with the relationship between \(\displaystyle H_{3}O^{+}\)

and \(\displaystyle OH^{-}\) concentration is:

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

Rearranging this equation gives:

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

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

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

The equilibrium governing the dissolution of \(\displaystyle CH_{3}COOH\) in water is:

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

\(\displaystyle CH_{3}COOH\) is the conjugate acid of \(\displaystyle CH_{3}COO^{-}\). In other words, \(\displaystyle CH_{3}COO^{-}\) is the conjugate base of \(\displaystyle CH_{3}COOH\).

Using the relationship, \(\displaystyle K_{w}=K_{a}K_{b}\), we can calculate the Kb.

By rearranging the equation we get:

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

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

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

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

Below is the equation to calculate pH:

\(\displaystyle pH=-log[H^{+}]\)

Therefore,

\(\displaystyle pH=-log[0.1M]\)

\(\displaystyle pH=1\)

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

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

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

Below is the equation to calculate pH:

\(\displaystyle pH=-log[H^{+}]\)

Therefore,

\(\displaystyle pH=-log[0.02M]\)

\(\displaystyle pH=1.7\)