Determine the pH of a solution prepared by mixing the following: 25.0 mL of 0.200 M HCl
75.0 mL of 0.100 M NaOH
50.0 mL of 0.200 M NaH₂PO₄
For H₃PO₄ K_a1 = 7.5 *10^-³, K_a2 = 6.2 * 10^-⁸, and K_a3 = 4.8 *10^-¹³.

Question

Quizplus · Accepted Answer

The correct answer is A because the pH of the solution is 6.73. The solution is acidic because the concentration of H+ ions is greater than the concentration of OH- ions. The H+ ions come from the HCl and the NaH2PO4. The OH- ions come from the NaOH. The pH of the solution can be calculated using the following equation:pH = -log[H+], where [H+] is the concentration of H+ ions in moles per liter.The concentration of H+ ions can be calculated using the following equation:[H+] = (moles of H+ ions) / (total volume of solution in liters)The moles of H+ ions can be calculated by adding the moles of H+ ions from the HCl and the NaH2PO4. The total volume of solution can be calculated by adding the volumes of the HCl, NaOH, and NaH2PO4 solutions.The moles of H+ ions from the HCl can be calculated using the following equation:moles of H+ ions from HCl = (concentration of HCl in moles per liter) * (volume of HCl in liters)The moles of H+ ions from the NaH2PO4 can be calculated using the following equation:moles of H+ ions from NaH2PO4 = (concentration of NaH2PO4 in moles per liter) * (volume of NaH2PO4 in liters) * (fraction of NaH2PO4 that dissociates)The fraction of NaH2PO4 that dissociates can be calculated using the following equation:fraction of NaH2PO4 that dissociates = K_a / (K_a + [H+])where K_a is the acid dissociation constant for NaH2PO4.The concentration of H+ ions in the solution can now be calculated by adding the moles of H+ ions from the HCl and the NaH2PO4 and dividing by the total volume of solution.The pH of the solution can now be calculated using the following equation:pH = -log[H+], where [H+] is the concentration of H+ ions in moles per liter.

Determine the PH of a Solution Prepared by Mixing the Following