A solution of formic acid (HCOOH, \(K_{\mathrm{a}}=1.8 \times 10^{-4}\) ) has a pH of \(2.70 .\) Calculate the initial concentration of formic acid in this solution.

The equilibrium expression for formic acid (HCOOH) dissociating into hydrogen ions (H+) and formate ions (HCOO-) is given by the equation: \[ \mathrm{K_a} = \frac{ [\mathrm{H^+}] [\mathrm{HCOO^-}]}{[\mathrm{HCOOH}]} \] We are given the value of Ka as \(1.8 \times 10^{-4}\). We will use this expression to solve for the initial concentration of formic acid later.

The pH of the solution is given as 2.70. Since pH is the negative logarithm of the hydrogen ion concentration, we can find the [H+] using the formula: \[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} \] Plugging in the pH value, we get: \[ [\mathrm{H^+}] = 10^{-2.70} \] Calculating the [H+] concentration, we get: \[ [\mathrm{H^+}] = 1.9953 \times 10^{-3} \mathrm{M} \]

The dissociation equation for formic acid is: \[ \mathrm{HCOOH} \leftrightarrows \mathrm{H^+} + \mathrm{HCOO^-} \] Since the formic acid dissociates into equal amounts of H+ and HCOO- ions, we can say that: \[ [\mathrm{HCOO}^-] = [\mathrm{H^+}] \] Now, using the calculated value for [H+], we find: \[ [\mathrm{HCOO}^-] = 1.9953 \times 10^{-3} \mathrm{M} \]

Now, let's assume the initial concentration of formic acid was C. At equilibrium, the concentration of HCOOH is (C - [HCOO-]). Using the Ka expression from step 1 and the known concentrations, we can write the equation: \[ \mathrm{K_a} = \frac{ [\mathrm{H^+}] [\mathrm{HCOO^-}]}{(C - [\mathrm{HCOO^-}])} \] Plugging in the given Ka value and calculated concentrations, we get: \[ 1.8 \times 10^{-4} = \frac{(1.9953 \times 10^{-3})(1.9953 \times 10^{-3})}{(C - 1.9953 \times 10^{-3})} \] Now, we can solve for the initial concentration of formic acid (C): \[ C = \frac{(1.9953 \times 10^{-3})^2}{1.8 \times 10^{-4}} + 1.9953 \times 10^{-3} = 0.022167 \mathrm{M} \] Therefore, the initial concentration of formic acid in the solution is approximately \(0.02217 \mathrm{M}\).