Question

Consider the titration of $80.0mLof0.100M Ba(OH{)}_{2}with 0.400 M HCl.$Calculate the pH of the resulting solution after each of the following volumes of HCl have been added.

1. $0.0 mL$
2. $20.0 mL$
3. $30.0 mL$
4. $40.0 mL$
5. $80.0 mL$
   

Short answer

1. The $\text{pH}$of the resulting solution is$13.30$ .

2. The $\text{pH}$of the resulting solution is$12.91$ .

3. The $\text{pH}$of the resulting solution is$12.56$ .

4. The $\text{pH}$of the resulting solution is 7.00.

5. The $\text{pH}$of the resulting solution is 1.30 .

   
   

   
   

Step-by-step solution

Definition of pH

A quantitative measure of the acidity or basicity of aqueous or other liquid solutions is defined as pH

Calculation of pH

1. Initial pH of the analyte solution
   $$\begin{gathered} \text{pOH}=0.70 \\ 14=\text{pH + pOH} \\ \text{14}-\text{0.70}=\text{pH} \\ \text{pH}=13.30 \end{gathered}$$

b. The pH of the solution is

$$\begin{gathered} \text{pOH}=1.09 \\ 14=\text{pH + pOH} \\ \text{14}-1.\text{09}=\text{pH} \\ \text{pH}=12.91 \end{gathered}$$

c. The pH of the solution is

$$\begin{gathered} \text{pOH}=1.44 \\ 14=\text{pH + pOH} \\ \text{14}-1.44=\text{pH} \\ \text{pH}=12.56 \end{gathered}$$

d.The reaction between barium hydroxide and hydrochloric acid is

${\text{Ba(OH)}}_{\text{2}}+\text{2HCl}\to {\text{BaCl}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}$

This means two moles of the acid react with one mole of the base.

There is no base remaining in the solution.The solution becomes neutral, and the pH is 7.00.

e.The total amount of base is $8\times {10}^{-3}\text{mol}$.

The amount of acid added to the baselocalid="1662225397892" $32\times {10}^{-3} \text{mol}$.

The reaction between barium hydroxide and hydrochloric acid is ${\text{Ba(OH)}}_{\text{2}}+\text{2HCl}\to {\text{BaCl}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}$

The amount of acid in the solution islocalid="1662225403819" $8\times {10}^{-3} \text{mol}$

The concentration of the acid is $0.05\text{mol/L}$

One mole of acid produces one mole of hydrogen ion.

$$\begin{gathered} \text{pH}=-\log [0.05] \\ \text{pH}=1.30 \end{gathered}$$