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Photon336

  • one year ago

Discussion/Questions acids and bases

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  1. Photon336
    • one year ago
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    370/371/372

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  2. Photon336
    • one year ago
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    373

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  3. Photon336
    • one year ago
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    @Rushwr @sweetburger @Woodward

  4. Rushwr
    • one year ago
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    370. I will go with C Beacuse Strong acids don't form conjugated acids. They form conjugate base . The conjugate base for this is NO3^-1

  5. Rushwr
    • one year ago
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    371. I'll go with A \[Ka = \frac{ [H ]^{2} [S ^{2-}]}{ [H _{2}S] }\]

  6. Rushwr
    • one year ago
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    this is the Ka equation for the overall reaction.

  7. Photon336
    • one year ago
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    i think it gives the Ka for the H2S

  8. Photon336
    • one year ago
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    Ka1 = 1x10^-7 ka just compares exponents so it would be i think 7

  9. Rushwr
    • one year ago
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    To find Ka we have o multiply Ka1 and Ka2 and get the pKa

  10. Photon336
    • one year ago
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    they say that because the second kA is so small you can just ignore it.

  11. Photon336
    • one year ago
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    but wait is that what you are supposed to do? i forgot that

  12. Photon336
    • one year ago
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    i'm just curious

  13. Rushwr
    • one year ago
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    IDK that but what I know is what I did . for 372 i'll go with D

  14. Rushwr
    • one year ago
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    wait it must be that then! What is the answer given for that is it 7?

  15. Rushwr
    • one year ago
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    373. it would be thymol blue !

  16. Rushwr
    • one year ago
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    Since it is a weak acid strong base titration the anionic hdrolysis occur making the medium basic at its equivalence point. Hence the pH increases !

  17. Photon336
    • one year ago
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    going to leave this open coming back later.

  18. Rushwr
    • one year ago
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    oki doki

  19. anonymous
    • one year ago
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    Well I thought 371 was supposed to be kind of like a trick question but then again I should probably review my acid base things. I thought we would say pKa of \(H_2S\) would be 7 and that the other value would be the value of the pKa of \(HS^-\).

  20. anonymous
    • one year ago
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    372 seems fun one cause you really get to compare pKas of all these conjugate acids and bases. I guess the question is, which one of these answer choices is most basic?

  21. Photon336
    • one year ago
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    @Woodward the solution said that because the second pKa was so small you could just ignore it.

  22. Photon336
    • one year ago
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    \[K _{3}PO _{4} ---> PO _{4} +3K ^{+}\] |dw:1439180017859:dw| \[PO _{4} + H2O --> HPO4 + OH ^{-}\] not sure

  23. Photon336
    • one year ago
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    372 not sure, think it would probably be the one that is most basic.

  24. Rushwr
    • one year ago
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    So the K+ ions are present in the solution right so won't it form KOH by combining with OH- ions. ?????? I am bad when it comes to inorganic. But what was the answer given? @Photon336

  25. Photon336
    • one year ago
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    yeah well.. let me see for 372. the answer is C.

  26. Photon336
    • one year ago
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    |dw:1439255147994:dw| |dw:1439255180440:dw| CH3CH2OH wouldn't raise the concentration of hydroxide ions, because it can't deprotinate water, don't think it's kb is that high. For basic compounds Kb > ka producing OH- ions the compound would have to act a as a base. and have a pKa that's higher than that of water, in -order to deprotinate it.

  27. Photon336
    • one year ago
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    @Rushwr it's poorly written but the PO4 ion produces the most OH- ion concentration. in this case i think that the PO4 ion is acting as a base to deprotinate water and produce OH ions. it's pKa must be higher than water i think in order to deprotinate the water. this would imply that it's not really stable in solution.

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