Answer
There will be precipitation.
Work Step by Step
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.
