Question 21M.2.hl.TZ2.5e

Alternative 1:
K_a = $\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HCOO}}^{-}\right]}{\left[\mathrm{HCOOH}\right]}$
OR
[HCOOH] = $\frac{{\left({10}^{-4.12}\right)}^2}{{10}^{-3.75}}$✓

«[HCOOH] =» 3.24 × 10⁻⁵ «mol dm⁻³» ✓

Alternative 2:
«pH = pK_a + log $\frac{\left[{\mathrm{HCOO}}^{-}\right]}{\left[\mathrm{HCOOH}\right]}$»
4.12 = 3.75 + log$\frac{{10}^{-4.12}}{\left[\mathrm{HCOOH}\right]}$ ✓

«[HCOOH] =» 3.24 × 10⁻⁵ «mol dm⁻³» ✓

Award [2] for correct final answer.