Rushwr
  • Rushwr
I have something for you!
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Rushwr
  • Rushwr
Rushwr
  • Rushwr
@vera_ewing this is for u cuz I thought u would like to go through some sooo I got all down from my teachr. Tell me if u can't read my letters.
Rushwr
  • Rushwr
@Photon336 hey cn u try the 5th sum for me? I am having a trouble with that. Can u also help me @vera_ewing

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Photon336
  • Photon336
For #5 you can get the number of mol of NAOH as well as acid because you know that at the stiochiometry point they are equal.
Photon336
  • Photon336
I'm not sure about the rest, let me tag someone I think @cuanchi might know
Cuanchi
  • Cuanchi
Yes @Photon336, she need the value of Ka for the lactic acid. According to wikipedia the pKa is 3.86 You need to calculate the concentration of the salt sodium lactate at the equivalence point. The number of salt moles is going to be equal to the number of moles of acid. The final volume is going to be the addition of the volume of the acid plus the volume of the base. Because both (the acid and the base) are the same concentration (0.112M) the volumes are going to be equal (25+25=50mL), then the final volume of the titration is going to be the double of the volume initial of the acid. If you have the same number of moles and the volume is double your concentration will be half of the initial concentration of the acid. Then the concentration of the salt at the equivalence point is going to be 0.112/2= 0.056M. You can do all this with calculations. (I can help you if you need). finally you have to calculate the pH at the equivalence point. This is a basic salt then the pH is going to be above 7. You have 3 options in your answers. You have to calculate. you can use an ICE table or just the shorcut formula [OH-]= sqr(( Kw/Ka) x [salt]) If you know [OH-] you easily can calculate pOH = -log [OH-] and then pH=14-pOH
Rushwr
  • Rushwr
Thank you so much @Cuanchi
Photon336
  • Photon336
So I guess you need to look up the Ka not asked to solve for it; going to work through this myself
Rushwr
  • Rushwr
@paki I want the answer for 5th question.
paki
  • paki
moles lactic acid = 0.02500 x 0.112 =0.0028 moles NaOH needed = 0.0028 volume NaOH = 0.0028 / 0.122 = 0.02295 L total volume = 0.02500 + 0.02295 = 0.0479 L [lactate] = 0.0028/ 0.0479 = 0.0584 M Now Find Ka; Ka = 1.38 x 10^-4 Kb = Kw/Ka =7.25 x 10^-11 = x^2 / 0.0834 - x x = [OH-]=2.46 x 10^-6 M pOH =5.61 pH =8.33 approx
Cuanchi
  • Cuanchi
I am sorry @Rushwr you didn't understand my explanation and you went to look just for the answer some were else. I am so disappointed!
Rushwr
  • Rushwr
NO No That is okai ! NO initially I understood and when i tried doing it again I was confused. I was looking who was online. since pakri was online I asked him. I am sorry for making u disappointed! I'm sorry again. I didn't mean it. I understood it clearly tht is why I closed the question also. Here I am really really sorry!
Rushwr
  • Rushwr
@Cuanchi
Cuanchi
  • Cuanchi
don't worry I feel better now
Rushwr
  • Rushwr
Hey are we good now?????? I'm sorry again! @Cuanchi
Cuanchi
  • Cuanchi
Yes sure just kidding with you a little. there are not so many question in this Summer time
Rushwr
  • Rushwr
phew I actually thought that u were pissed!

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