<u>Answer:</u> The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is
<u>Explanation:</u>
- <u>For 1:</u> At the beginning
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
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
![[H^+]=0.013M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.013M)
Putting values in above equation, we get:
To calculate the number of moles, we use the equation:
Molarity of nitric acid solution = 0.013 M
Volume of solution = 15 mL
Putting values in above equation, we get:
Molarity of methylamine solution = 0.017 M
Volume of solution = 10 mL
Putting values in above equation, we get:
- The chemical equation for the reaction of nitric acid and methylamine follows:
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 
So, 
of methyl amine will produce = 
To calculate the 
of base, we use the equation:
where,
= base dissociation constant = 
Putting values in above equation, we get:
- To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:
![pOH=pK_b+\log(\frac{[salt]}{[base]})](https://tex.z-dn.net/?f=pOH%3DpK_b%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%29)
![pOH=pK_b+\log(\frac{[CH_3NH_3^+]}{[CH_3NH_2]})](https://tex.z-dn.net/?f=pOH%3DpK_b%2B%5Clog%28%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%7D%7B%5BCH_3NH_2%5D%7D%29)
We are given:
![[CH_3NH_3^+]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BCH_3NH_3%5E%2B%5D%3D%5Cfrac%7B1.7%5Ctimes%2010%5E%7B-4%7D%7D%7B10%2B15%7D%3D6.8%5Ctimes%2010%5E%7B-6%7DM)
![[CH_3NH_2]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BCH_3NH_2%5D%3D%5Cfrac%7B1.7%5Ctimes%2010%5E%7B-4%7D%7D%7B10%2B15%7D%3D6.8%5Ctimes%2010%5E%7B-6%7DM)
Putting values in above equation, we get:
To calculate pH of the solution, we use the equation:
Hence, the pH of the solution is 4.59
