Question

How many grams of ammonium nitrate should be added to 250.00 mL of water to create a solution of pH=4.94?
Correct Answer:4.75 g

Answer 1

Very First,

we have to go in reverse direction

lets calculate the H+ ion concentration in a solution

pH=-log [H+]

4.94 = - log [H+]

[H+] = antilog[-4.96]

[H+] = 1.1482*10^-5

now when we add ammonium nitrate following reaction occurs

NH₄NO₃ = NH₄⁺ + NO3-

and NH₄⁺ reacts with water and gives H⁺ ion, acts as acid

NH₄⁺ + H₂O = NH^₃ + H₃O⁺

we know that Ka of NH4+ is 5.6 x 10^-10

Ka = [NH₃][H₃O⁺]/[NH₄⁺]

5.6 x 10^-10 = (1.1482*10^-5 )² / X(unknown)

X =(1.3184 *10^-10) / 5.6 x 10^-10

X = 0.2354 moles of NH₄⁺ per liter

we have to add ammonium nitrate not NH₄⁺ so we have to calculate the grams of ammonium nitrate

molecular weight of ammonium nitrate is 80.043 g/ mole

so for 0.2354 moles

0.2354g * 80.043 g/mole

=18.8444 grams of ammonium nitrate in 1 liter of solution

but our solution is 250 ml, means 1/4th of 1 liter

se we have to add 1/4th of 18.8444 gm

= 18.85/4

=4.74 grams of ammonium nitrate should be added to 250 ml of water

4.74gram is our answer