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 $HNO_3$.

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

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

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 $K_a$.

$K_a\times K_b=K_w$

$K_a=\frac{K_w}{K_b}=\frac{1\times 10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

Now we have to calculate the $pK_a$

$pK_a=-\log (5.55\times 10^{-10})$

$pK_a=9.25$

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

$pH=pK_a+\log \frac{[NH_3]}{[NH_4^+]}$

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

$pH=9.25+\log (\frac{0.001}{0.014})$

$pH=8.1$

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