Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) is a monoprotic acid that ionizes only partially in aqueous solutions. A \(0.10-M\) formic acid solution is \(4.2 \%\) ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of \(0.10 M\) formic acid.

We are given that the formic acid is \(4.2\%=0.042\) ionized in the solution. This value represents the degree of dissociation (α), and it is the fraction of acid molecules that dissociate into ions in the solution. The van't Hoff factor (i) can be calculated using the following formula for a monoprotic acid: \[ i = 1 - α \] In this case, we have: \[ i = 1 - 0.042 = 0.958 \]

We are given that the molarity and molality of the solution are the same, so the molality (m) is equal to the molarity, i.e., \(m = 0.10 \, \text{mol/kg}.\) For water, the molal freezing point depression constant (K_f) is \(1.86 \, ^\circ \text{C}/\text{molal}\), and the molal boiling point elevation constant (K_b) is \(0.512 \, ^\circ \text{C}/\text{molal}.\)

To calculate the freezing point depression (ΔT_f) and the boiling point elevation (ΔT_b), we can use the following formulas: \[ \Delta T_{f} = i \cdot K_{f} \cdot m \] \[ \Delta T_{b} = i \cdot K_{b} \cdot m \] Plugging in the values we found in Steps 1 and 2, we can find the values for ΔT_f and ΔT_b: \[ \Delta T_{f} = (0.958)(1.86 \, ^\circ \text{C}/\text{molal})(0.10 \, \text{mol/kg}) = 0.178 \, ^\circ \text{C} \] \[ \Delta T_{b} = (0.958)(0.512 \, ^\circ \text{C}/\text{molal})(0.10 \, \text{mol/kg}) = 0.049 \, ^\circ \text{C} \]

Now, we can calculate the new freezing point and boiling point of the formic acid solution. For water, the normal freezing point is \(0 ^\circ \text{C}\), and the normal boiling point is \(100 ^\circ \text{C}\). Therefore, the new freezing point and boiling point can be found using the following formulas: \[ \text{New Freezing Point} = \text{Normal Freezing Point} - \Delta T_{f} \] \[ \text{New Boiling Point} = \text{Normal Boiling Point} + \Delta T_{b} \] Plugging in the values we found in Steps 3: \[ \text{New Freezing Point} = 0 \, ^\circ \text{C} - 0.178 \, ^\circ \text{C} = -0.178 \, ^\circ \text{C} \] \[ \text{New Boiling Point} = 100 \, ^\circ \text{C} + 0.049 \, ^\circ \text{C} = 100.049 \, ^\circ \text{C} \] The freezing point of the \(0.10 M\) formic acid solution is approximately \(-0.178 \, ^\circ \text{C}\), and the boiling point of the solution is approximately \(100.049 \, ^\circ \text{C}\).