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JoannaBlackwelder

  • one year ago

In a titration of an unknown acid with .1M NaOH, we started with 25.5 mL of acid (with an intial pH of 2.5) and used 24.8 mL of the NaOH to get to the endpoint. What is the Ka of the acid?

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  1. JoannaBlackwelder
    • one year ago
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    @aaronq @abb0t @chmvijay

  2. JoannaBlackwelder
    • one year ago
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    @JFraser @taramgrant0543664 @wolfe8

  3. JoannaBlackwelder
    • one year ago
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    I got 1.03x10^-4, but that isn't an answer. Any thoughts?

  4. Photon336
    • one year ago
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    @cuanchi

  5. JoannaBlackwelder
    • one year ago
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    Thanks. I used a similar procedure to what you mentioned. I`m not sure why we get different answers though.

  6. cuanchi
    • one year ago
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    I got the same answer then @JoannaBlackwelder 1.03x10^-4 I dont understand why she said it is not the answer.What are the possible answers? Regarding to the @taramgrant0543664 answer I dont think is the correct answer. With that Ka the pH should be 2.06 or the concentration of the acid 0.013M. Please explain how did you got that answer.

  7. JoannaBlackwelder
    • one year ago
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    Ok, thanks @Cuanchi The given answers are 1.8x10^-5, 1.8x10^-4, and 1.3x10^-5 It is data collected from a lab, so that may have been the issue and not the calcs.

  8. cuanchi
    • one year ago
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    I will go for the 1.8x10^-4 with a theoretical pH of 2.4 the other two Ka will give you pH~ 2.8-2.9

  9. JoannaBlackwelder
    • one year ago
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    Yeah, that is what I was leaning toward too. Thanks so much!

  10. JFraser
    • one year ago
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    To find the Ka of the acid, you need to know its initial concentration, and it's pH, in order to fit into the equilibrium expression. I'd use stoichiometry to find the moles of acid neutralized, and the initial volume of acid will get you the starting concentration of the acid, [HA]. Knowing the pH will get you the staring concentration of both the [H+] and [A-] to plug into the equilibrium equation

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