Question

Desperately need help!

Answer 1

@SolomonZelman @TheSmartOne

Answer 2

2) use the relation between pH and pOH
pOH+pH=14
pH=14-pOH=14-(-log[OH-])
-log[H3O+]=14-(-log[OH-])

Answer 3

I got 2!!!!

Answer 4

I got 7.7 x 10^-13

Answer 5

I need more help with 3 than with 2 though if it wouldn't be too much trouble :/ I know how to do 5! :D

Answer 6

3) H3PO4-->H2PO4-+H+
Ka = [H+ ][ H2PO4-]/H3PO4
pH=-log[H+]
[H+]=5.011872336*10^-3 = [H2PO4-]
Ka=(5.011872336*10^-3)^2/8.6*10^-3

Answer 7

Need help with number 4!

Answer 8

set up the ICE box equation for the acid, and plug in the values

Answer 9

because it's the dissociation of and acid, the \(C_2H_3O_2^{-1}\) concentration will be \(equal\) to the \(H^{+1}\) concentration