## Chemistry: The Central Science (13th Edition)

1. Calculate $[AgIO_3]$ after the dissolution: $V_f = 10ml + 20ml = 30ml$ $C_i * V_i = C_f * V_f$ $0.01 * 20 = C_f * 30$ $C_f \approx 6.667 \times 10^{-3}M$ 2. Calculate $[NaIO_3]$ after the dissolution: $C_i * V_i = C_f * V_f$ $0.015 * 10 = C_f * 30$ $C_f = 5 \times 10^{-3}M$ 3. Find $[Ag^+]$ $[Ag^+] = [AgIO_3] \approx 6.667 \times 10^{-3}M$ 4. Find $[I{O_3}^-]$ $[I{O_3}^-] = [AgIO_3] + [NaIO_3]$ $[I{O_3}^-] = 6.667 \times 10^{-3} + 5 \times 10^{-3}$ $[{IO_3}^-] = 1.167 \times 10^{-2}M$ 5. Calculate the product of the concentrations for $AgIO_3:$ $P = [Ag^+][I{O_3}^-]$ $P = 6.667 \times 10^{-3} * 1.167 \times 10^{-2}$ $P = 7.780 \times 10^{-5}$ 6. Compare this value with the Ksp $Ksp = 3.1 \times 10^{-8}$ $P = 7.78 \times 10^{-5}$ Since P has a higher value, there will be precipitation.