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Rushwr

  • one year ago

I have something for you!

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  1. Rushwr
    • one year ago
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  2. Rushwr
    • one year ago
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    @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.

  3. Rushwr
    • one year ago
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    @Photon336 hey cn u try the 5th sum for me? I am having a trouble with that. Can u also help me @vera_ewing

  4. Photon336
    • one year ago
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    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.

  5. Photon336
    • one year ago
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    I'm not sure about the rest, let me tag someone I think @cuanchi might know

  6. cuanchi
    • one year ago
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    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

  7. Rushwr
    • one year ago
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    Thank you so much @Cuanchi

  8. Photon336
    • one year ago
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    So I guess you need to look up the Ka not asked to solve for it; going to work through this myself

  9. Rushwr
    • one year ago
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    @paki I want the answer for 5th question.

  10. paki
    • one year ago
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    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

  11. cuanchi
    • one year ago
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    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!

  12. Rushwr
    • one year ago
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    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!

  13. Rushwr
    • one year ago
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    @Cuanchi

  14. cuanchi
    • one year ago
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    don't worry I feel better now

  15. Rushwr
    • one year ago
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    Hey are we good now?????? I'm sorry again! @Cuanchi

  16. cuanchi
    • one year ago
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    Yes sure just kidding with you a little. there are not so many question in this Summer time

  17. Rushwr
    • one year ago
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    phew I actually thought that u were pissed!

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