Question

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Be sure to answer all parts. Calculate the molar solubility of Ca(I03)2 in each solution below. The Ksp of calcium iodate is 7.1 x 10-7 (a) 0.055 M Ca(NO3)2 (b) 0.055 M NaIO3

Answer 1

dus Concept & in both case kommen sons fresant in the solution that affects it Solubility Solution. a) 0.055M Ca(NO3)2 Dissociation Rep of Ca(NO3)2 Ca(NO3)2 (aq) Ca2+ (aq) + 2NO3 (ar) So. 0. 055m Ca(NO3)2 give ses 0.055M Cett Canticutration =) Ca (103) (3) - Cat(aq) +2 I 0 (aq) Initial. 0.055 0 ategben 0.055+0 2x Ksp = [cat] [I 0₃ J2 7. 18107 - (0.055+x) (2x) ' 7.1860172441 80 0.055+x 0.055 lo 701410-7 = (0.055) 462 x²2 71810-7 0.055X4 X = 5.68 810-3 so. Solubility of Ca (IO3)2 din solution is 5.68X10 3M

= no 055+21 aaaaaallllll b) 0.055 M NaIO Dissociation Ron of NQIO, NA I03 (5) = Mat (aq) +105 (02) So. 0.055 M NQIO; gives us the contiontzation of 10 is 0.055M = Ca (IO3), Cag) - Cart(aq) + 2 I 0 (0) O 0655 at ego. Ksp = [ca2+] [103]2 7.1810-7 - S X (0.055+238) :'7.18007 LULI 0.055+ 25 0.055 7017107 - S X(0.055)? s = 7.18107 (0.0552 S = 2.35 810-4 so solubility of cal I Os) is 2.35X10MM. Thank you, May it help you.