What is the pH of a solution that results when 0.010 mol HNO₃is added to 500.mL of a solution that is 0.10 M in aqueous ammonia and 0.55 M in ammonium nitrate.Assume no volume change.(The K_b for NH₃ = 1.8 $\times$ 10^-⁵. )

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Step 1: Write the balanced equation for the reaction between HNO₃ and NH3: HNO₃ + NH3 → NH4+ + NO3- Step 2: Write the expression for the equilibrium constant: K = [NH4+][NO3-]/[HNO₃][NH3] Step 3: Set up an ICE (initial, change, equilibrium) table: | | HNO₃ | NH3 | NH4+ (added) | NO3- (added) | |--------|----------------|------|--------------|--------------| | Initial | 0.010 M | 0.10 M | 0 M | 0 M | | Change | -x | -x | +x | +x | | Equil. | 0.010-x | 0.10-x | x | x | Step 4: Substitute equilibrium concentrations into the equilibrium constant expression and solve for x: 1.8 x 10^10 = (x)(x)/(0.010-x)(0.10-x) x = 6.3 x 10^-6 Step 5: Calculate [H+] using the concentration of NH4+: [H+] = 10^-pH = [NH4+] 0.55 M = [NH4+] pH = 8.40 Therefore, the pH of the solution is 8.40, and the correct answer is D.

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