Question

what is the pH of a solution that results when 0.010mol HNO3 is added to 500.ml of a solution that is 0.10M in aqueous ammonia and 0.55 M in ammonium nitrate. assume no volume change. (The Kb for NH3 =1.8 * 10-5 )

Answer 1

Answer : The  

pH of a solution is, 8.56

Explanation : Given,

$K_b=1.8\times 10^{-5}$

Concentration of ammonia (base) = 0.10 M

Concentration of ammonium nitrate (salt) = 0.55 M

First we have to calculate the value of $pK_b$.

The expression used for the calculation of $pK_b$ is,

$pK_b=-\log (K_b)$

Now put the value of $K_b$ in this expression, we get:

$pK_b=-\log (1.8\times 10^{-5})$

$pK_b=5-\log (1.8)$

$pK_b=4.7$

Now we have to calculate the pOH of buffer.

Using Henderson Hesselbach equation :

$pOH=pK_b+\log \frac{[Salt]}{[Base]}$

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

$pOH=4.7+\log (\frac{0.55}{0.10})$

$pOH=5.44$

The pOH of buffer is 5.44

Now we have to calculate the pH of a solution.

$pH+pOH=14\\\\pH+5.44=14\\\\pH=14-5.44\\\\pH=8.56$

Thus, the pH of a solution is, 8.56