Calculating the concentration of acetic acid in vinegar from pH
First off $$\ce{HC_2H_3O_2}\leftrightarrow \ce{H^+} + {\ce{C_2H_3O_2}}^-\\and\\ K_a = \frac{[\ce{H^+}][{\ce{C_2H_3O_2}}^-]}{[\ce{HC_2H_3O_2}]}$$ So let's do some simplifying. The amount of acetic acid left is the orignial molarity minus the amount that's dissociated. The amount of Hydrogen ion and conjugate base left is equal to the amount of that's been dissociated from the original amount of acid. So let's let HA equal the molarity of the acid and x equal to the amount dissociated. Soooooo $$K_a = \frac{[x][x]}{[HA-x]} = \frac{x^2}{HA-x}\\\text{Now let's solve}\\\rightarrow K_a(HA-x) = x^2\\\rightarrow HA -x= \frac{x^2}{K_a}\\\text{ switching the x interms of ph}\\ HA -10^{-ph} = \frac{(10^{-ph})^2}{K_a}\\\text{ plugging in}\\HA- 10^{-3} = \frac{(10^{-3})^2}{.0000174} = .0575$$ This is the molarity of the acid at equilibrium. Add 10^{-ph} for the concentration of the acid before dissociation.