Acids and Bases Question

Test Preparation AP Test Preparation
1

<p>Can somebody point me in the right direction for this question? I'm running in circles and I think it's really a lot easier than I'm making it...</p>

<p>In a different solution, 2.25 x 10^-4 mol of sodium benzoate (NaC6H5COO) and 3.25 x 10^4 mol of benzoic acid are dissolved in water to make 50.00 mL of solution.</p>

<p>What is the pH of this solution?</p>

2

<p>I think you're missing some information there. We need to know the Ka or Kb of the solution in order to find the answer. </p>

<p>Also, I think the benzoic acid isn't "10^4" mole because that would be too much, i think you meant to type 10^-4 moles.</p>

<p>Post the full question so we can help you better :)</p>

3

<p>You're right about the 10^-4 for benzoic acid... thanks :)</p>

<p>Okay, so I'm not completely lost - I thought there was a gap in the information I had too. My teacher gave us a worksheet with the 2006 Form B FRQ #1... and then this below it. In the FRQ, we didn't have to calculate Ka or Kb so I don't know what to do - that's exactly what's written on my homework.</p>

4

<p>Ha! I found the problem. You are missing the Ka value which is actually given on that question. </p>

<p>The 2006 ACTUAL form B FRQ #1 can be found here:
<a href="http://www.collegeboard.com/prod_downloads/student/testing/ap/chemistry/ap06_frq_chem_b.pdf%5B/url%5D"&gt;http://www.collegeboard.com/prod_downloads/student/testing/ap/chemistry/ap06_frq_chem_b.pdf&lt;/a&gt;&lt;/p>

<p>Cheers!</p>

5

<p>Wow... I feel incredibly stupid right now. (lol) Thanks for looking for me :)</p>

6

<p>Alright make an equilibrium table. If you don't know what this is, then ask your chem teacher. Some books also use this excellent method.</p>

<p>Sodium benzoate completely disassociates.</p>

<p>So in the table, initial moles:</p>

<p>Benz. acid:3.25x10^-4
benzoate 2.25x10^-4
hydornium:0</p>

<p>Benz. acid goes down X number of moles, while benzoate and hydronium go up X moles.</p>

<p>Final moles:
Benz. acid: (3.25x10^-4) -X
benzoate:(2.25x10^-4) +X
hydronium:X</p>

<p>setup equilibrium expression after converting to molarity. Do not divide the moles by liters to get molarity. Divide the real number (not X) by molarity. So for benzoic acid, the equilibrium molarity is 0.0065-X.</p>

<p>Ka=6.46x10^-5=(0.0045+X)X/(0.0065-x)</p>

<p>Solve for x.</p>

<p>x=9.02 x 10^-5 M= [H+]</p>

<p>-log[H+]= pH of 4.04</p>

<p>EDIT: IGNORE THAT SMILEY</p>