Photon336
  • Photon336
Discussion/Questions acids and bases
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Photon336
  • Photon336
370/371/372
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Photon336
  • Photon336
373
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Photon336
  • Photon336
@Rushwr @sweetburger @Woodward

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Rushwr
  • Rushwr
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
Rushwr
  • Rushwr
371. I'll go with A \[Ka = \frac{ [H ]^{2} [S ^{2-}]}{ [H _{2}S] }\]
Rushwr
  • Rushwr
this is the Ka equation for the overall reaction.
Photon336
  • Photon336
i think it gives the Ka for the H2S
Photon336
  • Photon336
Ka1 = 1x10^-7 ka just compares exponents so it would be i think 7
Rushwr
  • Rushwr
To find Ka we have o multiply Ka1 and Ka2 and get the pKa
Photon336
  • Photon336
they say that because the second kA is so small you can just ignore it.
Photon336
  • Photon336
but wait is that what you are supposed to do? i forgot that
Photon336
  • Photon336
i'm just curious
Rushwr
  • Rushwr
IDK that but what I know is what I did . for 372 i'll go with D
Rushwr
  • Rushwr
wait it must be that then! What is the answer given for that is it 7?
Rushwr
  • Rushwr
373. it would be thymol blue !
Rushwr
  • Rushwr
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 !
Photon336
  • Photon336
going to leave this open coming back later.
Rushwr
  • Rushwr
oki doki
anonymous
  • anonymous
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 \(HS^-\).
anonymous
  • anonymous
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?
Photon336
  • Photon336
@Woodward the solution said that because the second pKa was so small you could just ignore it.
Photon336
  • Photon336
\[K _{3}PO _{4} ---> PO _{4} +3K ^{+}\] |dw:1439180017859:dw| \[PO _{4} + H2O --> HPO4 + OH ^{-}\] not sure
Photon336
  • Photon336
372 not sure, think it would probably be the one that is most basic.
Rushwr
  • Rushwr
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
Photon336
  • Photon336
yeah well.. let me see for 372. the answer is C.
Photon336
  • Photon336
|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.
Photon336
  • Photon336
@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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