Question

Discussion/Questions acids and bases

Answer 1

370/371/372

Answer 2

373

Answer 3

@Rushwr @sweetburger @Woodward

Answer 4

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

Answer 5

371. I'll go with A
\[Ka = \frac{ [H ]^{2} [S ^{2-}]}{ [H _{2}S] }\]

Answer 6

this is the Ka equation for the overall reaction.

Answer 7

i think it gives the Ka for the H2S

Answer 8

Ka1 = 1x10^-7

ka just compares exponents so it would be i think 7

Answer 9

To find Ka we have o multiply Ka1 and Ka2 and get the pKa

Answer 10

they say that because the second kA is so small you can just ignore it.

Answer 11

but wait is that what you are supposed to do? i forgot that

Answer 12

i'm just curious

Answer 13

IDK that but what I know is what I did .
for 372 i'll go with D

Answer 14

wait it must be that then! What is the answer given for that is it 7?

Answer 15

373. it would be thymol blue !

Answer 16

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 !

Answer 17

going to leave this open coming back later.

Answer 18

oki doki

Answer 19

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 \(HS^-\).

Answer 20

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?

Answer 21

@Woodward the solution said that because the second pKa was so small you could just ignore it.

Answer 22

\[K _{3}PO _{4} ---> PO _{4} +3K ^{+}\]




\[PO _{4} + H2O --> HPO4 + OH ^{-}\]

not sure