Calculate the solubility of copper (I) iodate in 0.21 M copper (I) nitrate. Kap

Question

Calculate the solubility of copper (I) iodate in 0.21 M copper (I) nitrate. Kap is 7.4x10- M3 You should know that the Ksp must refer to the copper iodate because all nitrate compounds are soluble and strong electrolytes! Answer: Warm UP Question. What is the initial (before any reaction takes place) lead nitrate concentration when 9.0 mL of 0.237 M lead nitrate is added to 17 mL of 0.0061 M sodium chloride? Answer For the reaction: PbCl2(s)?Pb2+(aq)+2C11-(ag), what is Q* when 7.5 mL of 0.099 M lead nitrate is added to 14 mL of 0.025 M sodium chloride? Ksp of lead chloride is 1.6x105 M3 Hint given in general feedback Recall:Q is compared to Ksp to determine whether a precipitate forms. Answer: Remember each solution is diluted by the other solution. Sodium phosphate is added to a solution that contains 0.0062 M aluminum nitrate and 0.017 M calcium chloride. The concentration of the first ion to precipitate (either AlSor Ca2 decreases as its precipitate forms. What is the concentration of this ion when the second ion begins to precipitate? Answer

Explanation / Answer

Ans 1

The balanced reaction

Cu(IO3)2(s) = Cu2+(aq) + 2 IO3-(aq)

Let molar solubility of Cu(IO3)2 = x

molar solubility of IO3- = 2x

molar solubility of Cu2+ = Initial concentration of Copper nitrate + new concentration

= 0.21 + x   

Equilibrium constant expression of the reaction

Ksp = [Cu2+][IO3-]^2

7.4 * 10^-8 = (0.21+x) * (2x)^2

x << 0.21

7.4 * 10^-8 = 0.21 * ( 4x^2)

x = 2.97 * 10^-4 = molar solubility of Cu(IO3)2

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