What is the pH of the solution which results from mixing 50.0 mL of 0.30 M HF(aq)and 50.0 mL of 0.30 M NaOH(aq)at 25 $\circ$ C? (K_a of HF = 7.2 $\times$ 10^-4)

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Quizplus · Accepted Answer

The balanced equation for the reaction between HF and NaOH is: HF + NaOH → NaF + H2O We have equal moles of both HF and NaOH. So, the limiting reagent is the one with the fewer moles, which is either NaOH or HF depending on the amount of their moles initially. Assume NaOH is the limiting reagent in this case. Then, 50.0 mL of 0.30 M NaOH contains: 0.50 L × 0.30 mol/L = 0.15 mol NaOH According to the balanced equation, 0.15 mol NaOH reacts with 0.15 mol HF, which comes from 50.0 mL of 0.30 M HF. Therefore, the amount of unreacted HF is: 0.30 mol/L × (50.0 mL/1000 mL) - 0.15 mol = 0.015 mol The total volume of the solution is 100.0 mL, which is equal to 0.1 L. Thus, the concentration of HF in the solution is: 0.015 mol/0.1 L = 0.15 M The dissociation of HF in water is: HF + H2O ⇌ H3O+ + F- The equilibrium constant for this reaction, K_a, is given as 7.2 × 10-4 at 25°C. At equilibrium, the concentrations of the species can be expressed as: [H3O+] = [F-] = x M The concentration of HF in the solution is (0.30 - x) M, so: K_a = ([H3O+][F-])/([HF]) 7.2 × 10-4 = x2/(0.15 - x) Assuming x << 0.15, we can simplify this expression to: 7.2 × 10-4 = x2/0.15 x = √(7.2 × 10-4 × 0.15) = 0.021 M So, the pH of the solution is: pH = -log[H3O+] = -log(0.021) ≈ 1.98

What Is the PH of the Solution Which Results from Mixing