You wish to use precipitation to separate Br− ion from I− ions in a solution that is 0.50 M in each. You find Ksp values for the copper(I) salts are 6.3 x 10−9 for CuBr and 1.3 x 10−12 for CuI at 25°C. If you are required to reduce the value of one ion to 0.10% of its original value for “complete separation,” will selective precipitation by Cu+ work? Show your calculations and justify your answer.
You wish to use precipitation to separate Br− ion from I− ions in a solution that...
Copper(I) ions in aqueous solution react with NH3(aq) according to Cut(aq)2 NH (aq) - Cu(NH(aq) = 6.3 x 1010 Calculate the solubility (in g-L) of CuBr(s) (Ksp = 6.3 x 109) in 0.64 M NH3 (aq) solubility of CuBr(s): g/L
Copper(I) ions in aqueous solution react with NH3(aq) according to Cu+(aq)+2NH3(aq)⟶Cu(NH3)+2(aq)Kf=6.3×1010 Calculate the solubility (in g·L−1) of CuBr(s) (Ksp=6.3×10−9) in 0.27 M NH3(aq). solubility of CuBr(s):
Copper(I) ions in aqueous solution react with NH3(aq) according to Cu+(aq)+2NH3(aq)⟶Cu(NH3)+2(aq)?f=6.3×1010 Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) 2 + ( aq ) K f = 6.3 × 10 10 Calculate the solubility (in g·L−1) of CuBr(s) CuBr ( s ) ( ?sp=6.3×10−9 K sp = 6.3 × 10 − 9 ) in 0.50 0.50 M NH3(aq) NH 3 ( aq ) .
you have three flasks with 0.03 M Cl- , Br- I- if you have titrated each solution with Cu+ what order with the equivalence points appear in the titration curves? Ksp for CuI=1*10-12 Ksp for CuCl=1.9*10-7 Ksp for CuBr = 5 * 10-9 A. pCu+ (I-) < pCu+(Br-)<pCu+(Cl-) B.pCu+ (Cl-) < pCu+(Br-)<pCu+(I-) C.pCu+ (Br-) < pCu+(I-)<pCu+(Cl-)
Copper(1) ions in aqueous solution react with NH,(aq) according to Cu (aq) + 2NH, (aq) — Cu(NH) (aq) K = 6.3 x 100 Calculate the solubility (in g.L-') of CuBr(s) (Kp = 6.3 x 10-) in 0.10 M NH, (aq). solubility of CuBr(s):
Copper(I) ions in aqueous solution react with NH3(aq) according to Cu+(aq)+2NH3(aq)⟶Cu(NH3)2+(aq)Kf=6.3×1010 Calculate the solubility (in g·L−1) of CuBr(s) (Ksp=6.3×10−9) in 0.77 M NH3(aq).
Copper(I) ions in aqueous solution react with NH, (aq) according to Cut (aq) + 2 NH, (aq) + Cu(NH2) (aq) K = 6.3 x 1010 Calculate the solubility (in g.L-) of CuBr(s) (Kyp = 6.3 x 10') in 0.14 M NH3(aq). solubility of CuBr(s):
Copper(l) ions in aqueous solution react with NH3(aq) according to Cu(NH3);(aq) 10 Cu"(aq) +2NH3(aq) -> K,=6.3 x 10 Calculate the solubility (in g. L-1) of CuBr(s) (Ksp = 6.3x10-9) in 0.64 M NH3(aq). Number g/L
A solution contains 0.021 M Cl? and 0.017 M I?. A solution containing copper (I) ions is added to selectively precipitate one of the ions. At what concentration of copper (I) ion will a precipitate begin to form? What is the identity of the precipitate? Ksp(CuCl) = 1.0 × 10-6, Ksp(CuI) = 5.1 × 10-12. please show work! A) 4 .8 × 10-5 M, CuCl B) 3 .0 × 10-10 M, CuI C) 3 .0 × 10-10 M, CuCl D)...
Copper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f = 6.3 × 10 10 Calculate the solubility (in g·L−1) of CuBr ( s ) ( K sp = 6.3 × 10 − 9 ) in 0.68 M NH 3 ( aq ) . solubility of CuBr ( s )