Problem: What are the final hydrogen ion concentration and pH of a solution obtained by mixing 200 mL of 0.4 M aqueous NH3 with 300 mL of 0.2 M HCL? (KB = 1.8 X 10-5)
My first attempt at solving:
@ t0 I have 100% in the acid form (HCl)
mol HCl present = 0.3 L x 0.2 M = 0.06 mol HCl
mol HCl present = mol NH3 required
0.06 mol HCl = (x amount of NH3 in Liters) (0.4 M)
x L = 0.15 L = 150L of NH3(aq) :larrow:that doesn't make sense to me...
My 2nd attempt at solving:
pOH = pKb + log (molHCl / molNH3)
pOH = 4.74473 + log (.08/.06) = 4.86967
pH = 14 - pOH = 14 - 4.86967 = 9.13033
[H+] = 7.41E-10 M
The answer key states that the
[H+] = 1.66E-9 M and pH = 8.78
What am I doing wrong? Thank you in advance!!!
|All times are GMT. The time now is 11:38 AM.|
Powered by vBulletin® Version 3.8.4
Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
Copyright 2005 - 2012 Molecular Station | All Rights Reserved