What is the pH of a 0.001 M solution of acetylsalicylic acid (aspirin) (Ka 3.0 x 10-4)?

3 answers

Denote the chemical formula for the acid to be HA, then
Ka=[H+][A-]/[HA]=3.0*10-4
at a given temperature.

ACE table for
H+ A- HA
A 0 0 0.001
C x x -x
E x x 0.001-x

At equilibrium,
3.0*10-4=x²/(0.001-x)
Cross multiply to get
3.0*10-4(0.001-x)=x²
Solve for x, reject negative root.
and pH is -log10(x).
It should be between 3 and 3.5
(typo in format)
Cross multiply to get
3.0*10-4(0.001-x)=x²
Solve for x, reject negative root.
and pH is -log10(x).
It should be between 3 and 3.5.
Can't seem to get it right!
the -4 should be in superscript.