Question

gonjob PH: 4.74 + 'log 4x10-3 9. Find the pH of 60 mL of 0.8 M ammonia titrated with 100 mL of 0.6 M HCl. The Ko for ammonia is 1.8 x 10-5 bar and NH3 + HCl - NHA & Cl- cons-a - 4NH

Answer 1

Given:

M(HCl) = 0.6 M

V(HCl) = 100 mL

M(NH3) = 0.8 M

V(NH3) = 60 mL

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

mol(HCl) = 0.6 M * 100 mL = 60 mmol

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

mol(NH3) = 0.8 M * 60 mL = 48 mmol

We have:

mol(HCl) = 60 mmol

mol(NH3) = 48 mmol

48 mmol of both will react

excess HCl remaining = 12 mmol

Volume of Solution = 100 + 60 = 160 mL

[H+] = 12 mmol/160 mL = 0.075 M

use:

pH = -log [H+]

= -log (7.5*10^-2)

= 1.1249

Answer: 1.12