Respuesta :
Answer: The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is
Explanation:
- For 1: At the beginning
To calculate the pH of the solution, we use the equation:
[tex]pH=-\log[H^+][/tex]
We are given:
Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M
[tex][H^+]=0.013M[/tex]
Putting values in above equation, we get:
[tex]pH=-\log(0.013)\\\\pH=1.89[/tex]
- For 2:
To calculate the number of moles, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
- For nitric acid:
Molarity of nitric acid solution = 0.013 M
Volume of solution = 15 mL
Putting values in above equation, we get:
[tex]0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol[/tex]
- For methylamine:
Molarity of methylamine solution = 0.017 M
Volume of solution = 10 mL
Putting values in above equation, we get:
[tex]0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol[/tex]
- The chemical equation for the reaction of nitric acid and methylamine follows:
[tex]HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-[/tex]
As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.
By Stoichiometry of the reaction:
1 mole of methyl amine produces 1 mole of [tex]CH_3NH_3^+[/tex]
So, [tex]1.7\times 10^{-4}mol[/tex] of methyl amine will produce = [tex]\frac{1}{1}\times 1.7\times 10^{-4}=1.7\times 10^{-4}\text{ moles of }CH_3NH_3^+[/tex]
To calculate the [tex]pK_b[/tex] of base, we use the equation:
[tex]pK_b=-\log(K_b)[/tex]
where,
[tex]K_b[/tex] = base dissociation constant = [tex]3.9\times 10^{-10}[/tex]
Putting values in above equation, we get:
[tex]pK_b=-\log(3.9\time 10^{-10})\\\\pK_b=9.41[/tex]
- To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pOH=pK_b+\log(\frac{[salt]}{[base]})[/tex]
[tex]pOH=pK_b+\log(\frac{[CH_3NH_3^+]}{[CH_3NH_2]})[/tex]
We are given:
[tex]pK_b=9.41[/tex]
[tex][CH_3NH_3^+]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M[/tex]
[tex][CH_3NH_2]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M[/tex]
Putting values in above equation, we get:
[tex]pOH=9.41+\log(\frac{6.8\times 10^{-6}}{6.8\times 10^{-6}})\\\\pOH=9.41[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH=14-9.41=4.59[/tex]
Hence, the pH of the solution is 4.59
