Question

A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At the beginning (before base is added) pH = -log (H₂O) = -log(0.013) - 1-800 1.013) = 1.886 After adding 10.0 mL of C2H3NH2

Answer 1

Answer: The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

Step-by-step explanation:

• For 1: At the beginning

To calculate the pH of the solution, we use the equation:

`pH=-\log[H^+]`

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`

Putting values in above equation, we get:

`pH=-\log(0.013)\\\\pH=1.89`

• For 2:

To calculate the number of moles, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}`

• For nitric acid:

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

`0.013M=\frac{\text{Moles of }HNO_3* 1000}{15}\\\\\text{Moles of }HNO_3=1.95* 10^(-4)mol`

• For methylamine:

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

`0.017M=\frac{\text{Moles of }CH_3NH_2* 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7* 10^(-4)mol`

• The chemical equation for the reaction of nitric acid and methylamine follows:

`HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-`

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
`CH_3NH_3^+`

So,
`1.7* 10^(-4)mol` of methyl amine will produce =
`(1)/(1)* 1.7* 10^(-4)=1.7* 10^(-4)\text{ moles of }CH_3NH_3^+`

To calculate the
`pK_b` of base, we use the equation:

`pK_b=-\log(K_b)`

where,

`K_b` = base dissociation constant =
`3.9* 10^(-10)`

Putting values in above equation, we get:

`pK_b=-\log(3.9\time 10^(-10))\\\\pK_b=9.41`

• To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

`pOH=pK_b+\log(([salt])/([base]))`

`pOH=pK_b+\log(([CH_3NH_3^+])/([CH_3NH_2]))`

We are given:

`pK_b=9.41`

`[CH_3NH_3^+]=(1.7* 10^(-4))/(10+15)=6.8* 10^(-6)M`

`[CH_3NH_2]=(1.7* 10^(-4))/(10+15)=6.8* 10^(-6)M`

Putting values in above equation, we get:

`pOH=9.41+\log((6.8* 10^(-6))/(6.8* 10^(-6)))\\\\pOH=9.41`

To calculate pH of the solution, we use the equation:

`pH+pOH=14\\pH=14-9.41=4.59`

Hence, the pH of the solution is 4.59