please help within five minutes


label Chemistry
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

Acrylic acid is a weak monoprotic acid with Ka = 5.5×10-5 M. NaOH(s) was gradually added to 1.00 L of 8.58×10-2 M acrylic acid.

1)Calculate the pH of the solution after the addition of 1.20×10-1 mol of NaOH(s). 


Feb 16th, 2015

Please note that the concentration of H+, is:

[H+] = (8850-12v)/(1000+v) *10^-2 millimol/ml

where v is the volume of the NaOH added

using the value of your Equilibrium constant, I get:

v= 30.36 mL of  NaOH

so we get:

[H+]= 0.082 = 8.2 * 10^(-2)

so, we have:

pH = -log [H+] = 1.086


Feb 17th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Feb 16th, 2015
...
Feb 16th, 2015
Nov 25th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer