Question

Calculate the hydronium ion concentration and the pH when 15.0 ml. of 0.10 M NH, is mixed with 15.0 ml. of 0.10 M HCI (K = 5.6 x 10""). Concentration - pH- Sui An Try Another Version 9 ltem attempts remaining

Answer 1

Given:

M(HCl) = 0.1 M

V(HCl) = 15 mL

M(NH3) = 0.1 M

V(NH3) = 15 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 15 mL = 1.5 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 15 mL = 1.5 mmol

We have:

mol(HCl) = 1.5 mmol

mol(NH3) = 1.5 mmol

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

NH4+ here is strong acid

NH4+ formed = 1.5 mmol

Volume of Solution = 15 + 15 = 30 mL

Ka of NH4+ = 5.6*10^-10

concentration ofNH4+,c = 1.5 mmol/30 mL = 0.05 M

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

5*10^-2 0 0

5*10^-2-x x x

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

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.6*10^-10)*5*10^-2) = 5.292*10^-6

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

so, x = 5.292*10^-6 M

[H+] = x = 5.292*10^-6 M

use:

pH = -log [H+]

= -log (5.292*10^-6)

= 5.2764

Answer:

[H+] = 5.3*10^-6 M

pH = 5.28