(a) Write the net ionic equation for the reaction that occurs when a solution of hydrochloric acid (HCl) is mixed with a solution of sodium formate (NaCHO \(_{2} )\) . (b) Calculate the equilibrium constant for this reaction. (c) Calculate the equilibrium concentrations of \(\mathrm{Na}^{+}, \mathrm{Cl}^{-}, \mathrm{H}^{+}, \mathrm{CHO}_{2}^{-}\) and \(\mathrm{HCHO}_{2}\) when 50.0 \(\mathrm{mL}\) of 0.15 \(\mathrm{MCl}\) is mixed with 50.0 \(\mathrm{mL}\) of 0.15 \(\mathrm{MNaCHO}_{2} .\)

The balanced chemical equation for the reaction of HCl (a strong acid) with NaCHO2 (a salt containing a weak acid) can be written as: HCl(aq) + NaCHO2(aq) -> NaCl(aq) + HCHO2(aq)

Strong electrolytes like strong acids (HCl), strong bases, and soluble salts dissociate completely into ions in solution. So, we can write the net ionic equation for the reaction as: H+(aq) + CHO2^-(aq) -> HCHO2(aq)

The equilibrium constant, K, for any reaction can be written in terms of the concentrations of the products divided by the concentrations of the reactants, with each concentration raised to the power of the stoichiometric coefficients in the balanced net ionic equation: K = \(\frac{[HCHO2]}{[H^+][CHO2^-]}\) This reaction produces the conjugate acid (HCHO2) of the weak base (CHO2^-). Since the equilibrium constant expression is in terms of the weak acid dissociation constant (Ka), we can find K by using the relationship: K = \(\frac{1}{Ka}\) To find the value of Ka, use a reference for the Ka of the conjugate parent acid, formic acid (HCOOH or HCHO2): Ka = 1.8 x 10^-4 (for HCHO2) Now, we can find K: K = \(\frac{1}{1.8 \times 10^{-4}}\) = 5.56 x 10^3

The initial concentrations can be determined by using the volumes and molarities provided: Initial moles of HCl = (50.0 mL)(0.15 M) = 7.5 mmol Initial moles of NaCHO2 = (50.0 mL)(0.15 M) = 7.5 mmol The total volume of the solution is 100 mL. Initial concentration of H+ = \(\frac{7.5\,\text{mmol}}{100\,\text{mL}}\)= 0.075 M Initial concentration of CHO2- = \(\frac{7.5\,\text{mmol}}{100\,\text{mL}}\)= 0.075 M The reaction consumes equal moles of H+ and CHO2-, and produces equal moles of HCHO2: [H+] = 0.075 M - x [CHO2^-] = 0.075 M - x [HCHO2] = x

To find the equilibrium concentrations, we can plug the expressions for the concentrations into the equilibrium constant expression and solve for x: K = \(\frac{x}{(0.075-x)(0.075-x)}\) 5.56 x 10^3 = \(\frac{x}{(0.075-x)^2}\) Solving for x, we get about 0.0749 M. Now that we have x, we can find the equilibrium concentrations of each species: [Na+]=[H+] = 0.075 - 0.0749 ≈ 0.0001 M [Cl-] = 0.150 M (since the initial concentration of HCl was 0.15 M and the final concentration of H+ is about 0.0001 M) [H+] = 0.0001 M [CHO2^-] = 0.0001 M [HCHO2] = 0.0749 M Therefore, the equilibrium concentrations of Na+, Cl-, H+, CHO2^-, and HCHO2 are approximately 0.0001 M, 0.150 M, 0.0001 M, 0.0001 M, and 0.0749 M, respectively.