A student performed the lodate Analysis and found that it took 7.80 mL of 0.02 M

Question

A student performed the lodate Analysis and found that it took 7.80 mL of 0.02 M Na_2S_2O_3 to titrate 4.00 mL of saturated copper iodate. Answer the following questions: Using the balanced equations on page 79, what is the mole ratio of triiodide, I_3^-, to thiosulfate, S_2O_3^2-? What is the mole ratio of iodate. IO_3^- to triiodide, I_3"? Calculate the number of moles of iodate in 7.80 mL of 0.02 M Na_2S_2O_3. What is the iodate concentration in the 4.00 mL of saturated copper iodate solution? Use this value to find the K_p value for copper iodate?

Explanation / Answer

Important note -> if this is experimental, please add data

If this is theoretical, assume Ksp for Cu(IO3)2 (copper iodate) is given by Ksp = 6.94*10^–8 (many text reference have different values for Ksp, since those are pretty siensible)

now..

We know that

Cu(IO3)2(s) <--> Cu+2 + 2IO3-

Ksp = [Cu+2][IO3-]^2

so

[Cu+2] = 1 S

[IO3-] = 2S

substitute in Ksp

Ksp = [Cu+2][IO3-]^2

6.94*10^-8 = (S)(2S)^2

4*S^3 = 6.94*10^-8

S = ((6.94*10^-8)/4) ^(1/3) = 0.0025888 mol of Cu(IO3)2 per liter

we need IO3-2 so

[IO3-2] = 2*SW = 0.0025888*2 = 0.0051776M

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