Question

1a. If 315 mL of 1x10^-4M Ca(NO3)2 is mixed with 315 mL of 1 x10^-4M NaF, what will occur? For CaF2, Ksp=3.4x10^-11 (I know no precipitate forms but how do you do it)

2a. 2ClO2(aq)+2OH-(aq)-ClO3-(aq)+ClO2-(aq)+H2O(I)

Experiment number= 1,2 ,3
[ClO2](M) = 0.060,0.020,0.020
[OH-](M) =0.020,0.030,0.00276
Initial Rate(M/s)=0.0248,0.00276,0.00828

(A) What is the order of the overall reaction?

(B) What is the order of the reaction with respect to ClO2?

Answer 1

1. millimoles of Ca(NO3)2 = Molarity*Volume = 1*10^-4M * 315mL = 0.0315mmoles

so, millimoles of Ca = 0.0315mmoles

millimoles of NaF = 1*10^-4M*315mL = 0.0315mmoles

millimoles of F = 0.0315mmoles

Total Volume = 315+315 = 630mL

so, Concentration of Ca, [Ca] = 0.0315/630 M = 0.00005M

[F] = 0.0315/ 630 M = 0.00005M

CaF2 is formed when Ca(NO3)2 and NaF are mixed, so

Q = [Ca²⁺][F^-]²

= 0.00005*(0.00005)²

Q = 1.25*10^-13

Ksp = 3.4*10^-11

so, Q<Ksp no precipitate will form..