What is the solubility of cr(oh) at a ph of 11.10? (ksp cr(oh) is 6.70 10)?
Answers
Explanation:
pH+pOH=14
14-pH=pOH
pOH=14-11.6=2.4
pOH=-log[OH^-]
[OH^-]=10^-[2.4]
[OH^-]=3.98 x 10^-3 M= initial concentration
Cr(OH)3 -------> Cr^3+ + 3OH^-
Ksp=[x][3.98 x 10^-3 M]^3
Ksp/[3.98 x 10^-3 M]^3=x
x=6.7×10^-31/[3.98 x 10^-3 M]^3
Solve for x...
Hope further you will solve
Answer:
Use the solubility-product constant for Cr (OH) 3 ( Ksp = 6.7×10?31) and the formation constant for Cr (OH) ?4 from the following table to determine the concentration of Cr (OH) ?4 in a solution that is buffered at pH= 11 and is in equilibrium with solid Cr (OH) 3.
Table Formation Constants for Some Metal Complex Ions in Water at 25 ?C
Complex Ion Kf Equilibrium Equation
Ag(NH3)+2 1.7×107 Ag+(aq)+2NH3(aq)?Ag(NH3)+2(aq)
Ag(CN)?2 1×1021 Ag+(aq)+2CN?(aq)?Ag(CN)?2(aq)
Ag(S2O3)3?2 2.9×1013 Ag+(aq)+2S2O2?3(aq)?Ag(S2O3)3?2(aq)
CdBr2?4 5×103 Cd2+(aq)+4Br?(aq)?CdBr2?4(aq)
Cr(OH)?4 8×1029 Cr3+(aq)+4OH?(aq)?Cr(OH)?4(aq)
Co(SCN)2?4 1×103 Co2+(aq)+4SCN?(aq)?Co(SCN)2?4(aq)
Cu(NH3)2+4 5×1012 Cu2+(aq)+4NH3(aq)?Cu(NH3)2+4(aq