[soupastupid]

This is waht im thinking

NH4NO3 is neutral because
NH4 is conj acid of weak base NH3
NO3 is conj base of strong acid HNO3

HCN is a weak acid

and so the only contributor to pH is NO3
waht do i do from here?

[soupastupid]

do i just take

-log(.26)

and get .59 for pH?

[Borek]

NH4 is conj acid of weak base NH3

And - as such - what acid it is? WHat pKa does it have?

NO3 is conj base of strong acid HNO3

Strong base? Weak base?

HCN is a weak acid

Correct. Look pKa in tables.

and so the only contributor to pH is NO3

Quite the opposite.

What acids does this solution contains? Will much weaker acid dissociate in the presence of the much stronger acid? Note, that by much stronger I don't mean generally strong, just relatively stronger.

[soupastupid]

NH4 is a weak acid with pKA of 9.25
NO3 is neutral
HCN is a weak acid with pKA 9.3

HCN is stronger than NH4 (barely)
weaker acid does dissociate with stronger acid present but is not a major contributor to pH

so i get

HCN <--> H+ + CN
.105          0       0
+x             -x      -x

HCN ka = 4.9*10^-10 = (x^2)/(.105)

x= 7.17*10^-6
pH = 5.14

[Borek]

NH4 is a weak acid with pKA of 9.25

Wrong. Think it over.

And please use charges when listing ions.

[soupastupid]

this right?

[AWK]

NH4 is a weak acid with pKA of 9.25
NO3 is neutral
HCN is a weak acid with pKA 9.3

HCN is stronger than NH4 (barely)
weaker acid does dissociate with stronger acid present but is not a major contributor to pH

so i get

HCN <--> H+ + CN
.105          0       0
+x             -x      -x

HCN ka = 4.9*10^-10 = (x^2)/(.105)

x= 7.17*10^-6
pH = 5.14

Both weak acids contribute to pH. You can eventually neglect a contribution of this acid which shows the products K_a x c at least 100 times lower.
The general approximated formula is
[H⁺]= SQRT(K_a1 x c₁ + K_a2 x c₂)

[soupastupid]

this right?

[Borek]

Close, my result differ by about 0.1 unit. No idea what and why you did with HCN, but it is wrong.

[soupastupid]

with HCN this is what i did

it is the weaker acid and so

after i get the H3O conc from NH4

1.21*10^-5

i plug that into the HCN

amd so i get

HCN <--> H+ + CN-
Initial CONC .105        1.21*10^-5  and zero for CN- conc
                   -x               +x                 +x

ka= 4.9*10^-10= (1.21*10^-5)x + x^2
                                .105
   x^2+(1.21*10^-5)x - 5.145*10^-11   = 0

QUADRATIC

X= 3.33*10^-6
H3O+ = (1.21*10^-5) + 3.33*10^-6 = 1.54*10^-5
pH= -log [H3O+]  = 4.81

what am i doing wrong??

[Borek]

Sorry, my mistake. Looks like your approach is correct.

Please comment on what you are doing, at the moment you force everyone to guess what you did - it is a waste of time. And if we guess wrong, you waste both our time and your own.

[soupastupid]

i dont kno wats wrong

the quiz said the answer was 4.56

the quiz is wrong huh?

[Borek]

4.56 looks completely off to me.

See attachement, these are exact concentrations calculated for this system. While 4.85 is not the same as your 4.81, it is close enough - and the difference is the effect of approximations used. 4.56 looks way off.

[soupastupid]

!!

i hate that stupid quiz
i never know if its me or the quiz

i spent hours trying to figure out what the **** i was doing wrong.

i guess its good to know that i know how to do the problem.

i must have done that problem 50 times man

thanks for your *delete me*

how did u get that program you're using to calculate pH?
should i get that thing?

[enahs]

I am sure he said you should, as it is his software!
Checkout the website in his signature for the software. As well as the "lectures" / tutorials on his site for calculations are pretty much the best on the internet.