Solubility Homework Set Worked Out Solutions

# 76x10 cdco3s4nh3aqcdnh342aqco32aq 15m 0m 0m 4xm xm xm

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: OH‐ is 2x: [OH‐]=2x=2(4.0 x 10‐7 M)= 8.0 x 10‐7 M [H+][OH‐]=1.0 x 10‐14 +](8.0 x 10‐7 M )=1.0 x 10‐14 [H [H+]=1.2 x 10‐8 M pH=‐log[H+]=‐log(1.2 x 10‐8)=7.9 33. What is the molar solubility (mol/L) of Cr(OH)3 at pH = 10.00? Ksp for Cr(OH)3 is 6.3 x 10–31 and Kf for Cr(OH)4– is 8 x 1029. In this problem, a complex ion is formed. This effect increases solubility and trumps both the common­ion and pH effects. To determine the molar solubility of a solid whose ion forms a complex ion, we need to find the overall equilibrium expression and “overall K” for both the dissolution of the solid and the formation of the complex‐ion. Identical species on opposite sides of the equilibrium arrow cancel. When “adding” two equilibria, the overall K is the product of the individual K values: Cr3+(aq) + 3OH‐(aq) Ksp=6.3 x 10‐31 Cr(OH)3(s) 3+(aq) + 4OH‐(aq) ‐(aq) Cr(OH4) Kf=8.0 x 1029 Cr ‐(aq) ‐(aq) Cr(OH)3(s) + OH Cr(OH4) K= Ksp x Kf ‐31 29 =(6.3 x 10 )(8.0 x 10 ) =5.04 x 10‐1 ‐ from the given pH: Calculate the initial concentration of OH 8 Dr. Fus CHEM 123 pH=‐log[H+]=10.00 [H+]=1.0 x 10‐10 M +][OH‐]=1.0 x 10‐14 (always true) [H (1.0 x 10‐10 M) +][OH‐]=1.0 x 10‐14 [OH‐]=1.0 x 10‐4 M Now, set up an ICE table for the overall expression using the overall K. Assuming the solution is buffered at a pH of 10.00, the concentration of OH‐ will not change: Cr(OH)3(s) + OH‐(aq) Cr(OH4)‐(aq) Initial ‐‐‐‐‐ 1.0 x 10‐4 M 0 M Change ‐‐‐‐‐ NO CHANGE +x M Equilibrium ‐‐‐‐‐ 1.0 x 10‐4 M x M K=[Cr(OH4)‐]= ___x____ = 5.04 x 10‐1 [OH‐] 1.0 x 10‐4 x= 5.0 x 10‐5 M 34. Calculate the concentration of free cadmium ion, [Cd2+], in a solution that contains 0.20 M Cd(NO3)2 in 2.0 M NaCN. Cadmium ion forms the complex ion, Cd(CN)42‐ for which Kf is 6.0 x 1018. To solve problems like this, you need to take a 2‐step approach. First, assume all of the Cd2+ is converted to the complex ion: Cd2+(aq) + 4CN‐(aq) Cd(CN)42‐(aq) Initial 0.20 M 2.0 M 0 M Change ‐0.20 M ‐4(0.20) M +0.20 M Final (NOT equilibrium) 0 M 1.2 M 0.20 M Next, use the FINAL concentrations from the first table as the INITIAL concentrations for a new table of the REVERSE reaction (dissociation of the complex ion) to determine the equilibrium concentration of free Cd2+: Cd2+(aq) + 4CN‐(aq) Cd(CN)42‐(aq) Initial 0.20 M 0 M 1.2 M Change ‐x M +x M +4x M Equilibrium 0.20‐x M xM 1.2+4x M These equilibrium concentrations refer to the reverse reaction, so they do not correspond to Kf, instead they correspond to Kd, the dissociation constant, which is 1/Kf (K for a reverse reaction always equals the inverse of K for the forward reaction): Kd=[Cd2+][CN‐]4= (x)(1.2+4x)4 = _2.07x_=___1_____ [Cd(CN)42‐] (0.20‐x) 0.20 6.0 x 1018 x= 1.6 x 10‐20 M 9 Dr. Fus CHEM 123 35. Calculate the molar solubility of CdCO3 in 1.5 M NH3. Note that Cd2+ forms the Cd(NH3)42+ complex ion for which Kf is 1.3 x 107. Ksp for CdCO3 is 5.2 x 10‐12. Init...
View Full Document

## This homework help was uploaded on 03/28/2014 for the course CHEM 123 taught by Professor Larosa during the Spring '06 term at Ohio State.

Ask a homework question - tutors are online