What is the pH of a solution prepared by mixing 48.00 mL of 0.25 M H₃PO₄ with 85.00 mL of 0.85M LiH₂PO₄? Assume that the volume of the solutions are additive and that K_a = 7.1 × 10^-3 for H₃PO₄. A) 11.07 B) 2.93 C) 4.31 D) 8.63

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First, calculate the moles of each acid: moles H₃PO₄ = (0.25 M) x (0.04800 L) = 0.012 moles moles LiH₂PO₄ = (0.85 M) x (0.08500 L) = 0.07225 moles Now, calculate the total moles and concentrations of each ion: moles H+ = moles H₃PO₄ + moles LiH₂PO₄ = 0.08425 moles total volume = 48 mL + 85 mL = 133 mL = 0.133 L [H+] = moles H+ / total volume = 0.08425 moles / 0.133 L = 0.634 M Now, use the Ka expression for H₃PO₄ to find the pH: Ka = [H+][₂PO4₄-] / [H₃PO4₄] 7.1 x 10^-3 = (0.634 M)(x) / (0.012 moles) x = 1.34 x 10^-11 pH = -log[H+] = -log(1.34 x 10^-11) = 2.93

What Is the PH of a Solution Prepared by Mixing