Determine the pH of the following solution. Initial concentrations are given. [NaCHO₂] = 0.815 M, [HBr] = 0.105 M, Ka (HCHO₂) = 1.8 × 10^-4 A) 2.91 B) 4.63 C) 4.27 D) 9.43 E) 11.09

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First, we need to write the equation for the dissociation of HCHO₂: HCHO₂ + H2O ⇌ CHO₂- + H3O+ The Ka value tells us that this is a weak acid, so we can use the following approximation: [H3O+] = √(Ka *[HCHO₂] / [CHO₂-]) Next, we need to calculate the concentration of CHO₂- using the initial concentration of NaCHO₂. Since NaCHO₂ is a salt of the weak acid, we can assume that it completely dissociates in water: NaCHO₂ → Na+ + CHO₂- [CHO₂-] = [NaCHO₂] = 0.815 M Now we can use the equation for the equilibrium constant of water to solve for [H3O+]: Kw = [H3O+][OH-] = 1.0 × 10^-14 at 25°C Assuming that [OH-] is very small compared to [H3O+], we can approximate this as: [H3O+] ≈ Kw / [OH-] Since we know that the solution is acidic (due to the presence of HBr), we can assume that [H3O+] > [OH-]. Therefore: [H3O+] ≈ Kw / [OH-] > Kw / [H3O+] [H3O+]^2 > Kw [H3O+] > √Kw = 1.0 × 10^-7 M Finally, we can use the equation for pH: pH = -log[H3O+] Substituting in our values, we get: [H3O+] = √(Ka *[HCHO₂] / [CHO₂-]) = √(1.8 × 10^-4 * 0.105 / 0.815) ≈ 1.41 × 10^-3 M pH = -log(1.41 × 10^-3) ≈ 4.27

Determine the PH of the Following Solution