[MathyMed]

This question was in Osequim 2018 (Chemistry Olympics of Sergipe) for 2nd graders in high school (about 15/16 years old) and it's not normal to students learn organic chemistry at this grade, however this question appeared at the test and me as well as everybody else, got 0 points from this question, as you can see in these pictures: https://prnt.sc/u799en
https://prnt.sc/u79dly

(Translation)
http://prntscr.com/u79cjb
Question 12: Sulphanilic acid, used in the manufacture of dyes, is prepared by the reaction of aniline with acid
sulfuric: (see picture)
The acid has a pKa value of 3.23. The sodium salt of the acid, Na (H2NC6H4SO3), is very soluble in water.
If you dissolve 1.25 g of the salt in 125 mL of solution, what is the pH of the solution?

I've been looking for help to solve this question since the day of the test (May 2018) and since then i still have no clue how to solve this

[AWK]

Can you solve the problem: what is pH 0.05 M sodium acetate (acetic acid has a pKa of 4.77)?

[Anduril]

Do you happen to know what answer is correct? It would help

[AWK]

For my question answer is pH=11.0 (10.96).
For Brasilian question, I got the result by 0.01 units lower from that given by OLimpiad Organizers (this difference is unimportant).

[MNIO]

I suppose I'd do this.
*********
after dissolving the salt in solution NaA --> Na(+) + A(-)   you then have this reaction
  A(-) + H2O <--> HA + OH-
with
  kb = [HA][OH-]/[A-] = kw / ka = 1x10-14 / 10-3.23 = 1.698x10^-11

initial concentration of A-
  (1.25g / 0.125L) * (1 mol / 172.15g) = 0.05809 M

from an ICE table
                            [A-]         [HA]       [OH-]
  initial               0.05809        0             0
  change                 -x            +x           +x
  equilibrium      0.05809-x      x              x

plugging that back into the equation for Kb
  x2 / (0.05809 - x) = 1.698x10^-11
  x = 9.932x10^-7 = [OH-]

so that
  pOH = -log(9.932x10^-7) = 6.00
  pH = 14.00 - 6.00 = 8.00

[AWK]

The molecular mass of sodium sulfanilate C₆H₆NNaO₃S is 195.171.

[MNIO]

is it?  ooops... my bad. I looked at the (H2NC6H4SO3) and forgot to add the sodium.. ok then.. recalculated

*********
same to here.  then

initial concentration of A-
  (1.25g / 0.125L) * (1 mol / 195.17g) = 0.05124 M

from an ICE table
                            [A-]         [HA]       [OH-]
  initial               0.05124        0             0
  change                 -x            +x           +x
  equilibrium      0.05124-x      x              x

plugging that back into the equation for Kb
  x2 / (0.05124 - x) = 1.698x10^-11
  x = 9.328x10^-7 = [OH-]

so that
  pOH = -log(9.328x10^-7) = 6.03
  pH = 14.00 - 6.00 = 7.97

[AWK]

is it?  ooops... my bad. I looked at the (H2NC6H4SO3)  ok then.. recalculated


  pH = 14.00 - 6.00 = 7.97

Copy/paste error.