Question

In a titration of 46.24 mL of 0.3359 M ammonia with 0.3359 M aqueous nitric acid, what is the pH of the solution when 46.24 mL of the acid have been added?

Answer 1

we have:

Molarity of HNO3 = 0.3359 M

Volume of HNO3 = 46.24 mL

Molarity of NH3 = 0.3359 M

Volume of NH3 = 46.24 mL

mol of HNO3 = Molarity of HNO3 * Volume of HNO3

mol of HNO3 = 0.3359 M * 46.24 mL = 15.532 mmol

mol of NH3 = Molarity of NH3 * Volume of NH3

mol of NH3 = 0.3359 M * 46.24 mL = 15.532 mmol

We have:

mol of HNO3 = 15.532 mmol

mol of NH3 = 15.532 mmol

15.532 mmol of both will react to form NH4+ and H2O

NH4+ here is strong acid

NH4+ formed = 15.532 mmol

Volume of Solution = 46.24 + 46.24 = 92.48 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10

concentration ofNH4+,c = 15.532 mmol/92.48 mL = 0.1679 M

NH4+ + H2O -----> NH3 + H+

0.1679 0 0

0.1679-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*0.1679) = 9.659*10^-6

since c is much greater than x, our assumption is correct

so, x = 9.659*10^-6 M

[H+] = x = 9.659*10^-6 M

we have below equation to be used:

pH = -log [H+]

= -log (9.659*10^-6)

= 5.01

Answer: 5.01