The equivalence point in a titration is the point in which the number of moles of titrant equals to the number of moles of analyte:

\(\displaystyle number\ of\ moles\ of\ acetic\ acid=number\ of\ moles\ NaOH\)

We need to determine the number of moles of acetic acid we are dealing with:

\(\displaystyle 50.00\ mL=0.050\ L\)

\(\displaystyle \frac{0.150\ moles}{L}\times0.050L\ acetic\ acid=0.0075\ moles\ acetic\ acid\)

Therefore, at the equivalence point there are also 0.0075 moles of \(\displaystyle NaOH\) in the solution. In order to determine the number of moles of \(\displaystyle NaOH\) in solution, we must use the concentration of \(\displaystyle NaOH\) as a conversion factor to determine the volume added to the acetic acid solution.

\(\displaystyle 0.0075\ moles\ NaOH\times \frac{L}{0.100\ moles}=0.075\ L\ NaOH\)

A pH indicator added to a solution causes a color change which is an indication that a reaction has reached the equivalence point. The equivalence point is when the number of moles of titrant equals the number of moles of analyte. The end point is an approximation of this in a titration.

At the equivalence point:

\(\displaystyle moles\ of\ NaOH=moles\ of\ acid\)

The number of moles of NaOH at equivalence point is calculated as follows:

\(\displaystyle 0.030\ L\times \frac{0.05\ moles}{L}=0.0015\ moles\ NaOH\)

\(\displaystyle moles\ of\ acid= 0.0015\ moles\)

Assuming we were trying to convert the number of grams of the monoprotic acid to moles, the equation would be set up as follows:

\(\displaystyle grams\ of\ acid\ (g) \times\ molecular\ weight\left ( \frac{moles}{g} \right )=number\ of\ moles\)

Let's plug the values we have into this equation and give the molar mass a value of \(\displaystyle X\):

\(\displaystyle 0.4000\ g\times \frac{1\ mole}{X}=0.0015\ moles\)

Rearranging this equation to solve for the molar mass \(\displaystyle X\) gives:

\(\displaystyle X=\frac{0.4000\ g\times\ 1\ mole}{0.0015\ moles}=266.67\ grams\)

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

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 \(\displaystyle HCl\).

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

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

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