Question

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 28.0 mL of HNO3.

Answer 1

A) Initial moles of NH3 = 75/1000 x 0.200 = 0.015 mol

Moles of HNO3 added = 28/1000 x 0.500 = 0.014 mol

NH3 + HNO3 => NH4+ + NO3-

Moles of NH3 left = 0.015 - 0.014 = 0.001 mol

Moles of NH4+ = 0.014 mol

Ka(NH4+) = Kw/Kb(NH3)

= 10-14/1.8 x 10-5 = 5.556 x 10-10

Henderson-Hasselbalch equation:

pH = pKa + log([NH3]/[NH4+])

= -log Ka + log(moles of NH3/moles of NH4+) since volume is the same for both

= -log(5.556 x 10-10) + log(0.0065/0.014)

= 8.14

Answer 2

Answer : The pH after the addition of 28.0 ml of
`HNO_3` is, 8.1

Explanation :

First we have to calculate the moles of
`NH_3` and
`HNO_3`.

`\text{Moles of }NH_3=\text{Concentration of }NH_3* \text{Volume of solution}=0.200M* 0.075L=0.015mole`

`\text{Moles of }HNO_3=\text{Concentration of }HNO_3* \text{Volume of solution}=0.500M* 0.028L=0.014mole`

The balanced chemical reaction is,

`NH_3+HNO_3\rightarrow NH_4^++NO_3^-`

Moles of
`NH_3` left = Initial moles of
`NH_3` - Moles of
`HNO_3` added

Moles of
`NH_3` left = 0.015 - 0.014 = 0.001 mole

Moles of
`NH_4^+` = 0.014 mole

Now we have to calculate the
`K_a`.

`K_a* K_b=K_w`

`K_a=(K_w)/(K_b)=(1* 10^(-14))/(1.8* 10^(-5))=5.55* 10^(-10)`

Now we have to calculate the
`pK_a`

`pK_a=-\log (5.55* 10^(-10))`

`pK_a=9.25`

Now we have to calculate the pH by using Henderson-Hasselbalch equation.

`pH=pK_a+\log ([NH_3])/([NH_4^+])`

Now put all the given values in this expression, we get:

`pH=9.25+\log ((0.001)/(0.014))`

`pH=8.1`

Therefore, the pH after the addition of 28.0 ml of
`HNO_3` is, 8.1