Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
False. Because gas to a liquid is called condensation
The four ionic species initially in solution are Na⁺, PO₄³⁻, Cr³⁺, and Cl⁻. Since the precipitate is composed of Cr³⁺ and PO₄³⁻ ions, the spectator ions must be Na⁺ and Cl⁻.
The complete ionic equation is 3Na⁺(aq) + PO₄³⁻(aq) + Cr₃⁺(aq) + 3Cl⁻(aq) → 3Na⁺(aq) + 3Cl⁻(aq) + CrPO₄(s).
So the balanced <u>net ionic equation</u> for this reaction would be Cr³⁺(aq) + PO₄³⁻(aq) → CrPO₄(s).
