what is Will SrC_2O_4 precipitate when 0.25 g of K_2C_2O_4 are added to 1 L of a

Question

what is Will SrC_2O_4 precipitate when 0.25 g of K_2C_2O_4 are added to 1 L of a 0.05 M solution of Sr^+2? Show work to justify your conclusion. What is the maximum mass of K_2CO_3 that can be added to 1 L of 0.01 M Zn^+2 without precipitating ZnCO_3? You have 1 L of a solution containing 0.01 M of both Mg^+2 and Ca^+2 ions. You slowly add K_2CO_3. Which precipitates first, CaCO_3, or MgCO_3? Why? At what CO_3^-2 concentration does this molecule precipitate? What mass of K_2CO_3? At what CO_3^-2 concentration does the second species, CaCO_3, or MgCO_3, precipitate? What is the concentration of the first ion to precipitate, Mg^+2 or Ca^+2, when the second precipitates?

Explanation / Answer

Q3.

SrC2O4

m = 0.25 g of K2C2O4 added to

0.05 M of Sr+2

SrC2O4(s) <--> Sr+2(aq) + C2O4-2(aq)

Ksp = [Sr+2][C2O4-2]

5*10^-8 = [Sr+2][C2O4-2]

get

[Sr+2] = 0.05 M

[C2O4-2] = mol of C2O4-2 / 1 L

mol of K2C2O4 = mass/MW = 0.25/166.2156= 0.00150 mol of C2O4-2

so

Q = [Sr+2][C2O4-2] =0.05* 0.00150 = 0.000075 = 7.5*10^-5

since

Q > Ksp

expect precipitate

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