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Including activity coefficient, find the concentration of Ba2+in 0.100M(CH3)4NIO3solution saturated with Ba(IO3)2.

Short Answer

Expert verified

The concentration of Ba2+in a 0.001MCH34NIO3solution saturated with BaIO32is6.610-7M

Step by step solution

01

Step 1:Finding the activity coefficient of aqueous solution Ba2+ and IO3-.

In the given problem, we need to find the activity coefficient along with concentration of Ba2+in a 0.001MCH34NIO3solution saturated with BaIO32.

The solubility of BaIO32is not quite expressed. So we will assume that BaIO32contributes negligible IO3-to0.100MCH34NIO3.

cIO3-=0.1M

From the Table 8-1, we can know that the values are,

Ba2+=0.380IO3-=0.775

02

Finding the concentration of  Ba2+in a 0.001M(CH3)4NIO3 solution saturated with  Ba(IO3)2 using equilibrium constant equation.

Now, we will use equilibrium constant equation,

Ksp=1.510-9Ksp=cBa2+(Ba2+)c2IO3-1.510-9=cBa2+0.3800.120.7752cBa2+=6.610-7M

Thus the concentration of Ba2+in a 0.001MCH34NIO3solution saturated with BaIO32is 6.610-7M

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Most popular questions from this chapter

Using activities, calculate the pH of a solution containing 0.010 M NaOH plus 0.012 0 M LiNO3 . What would be the pH if you neglected activities?

Systematic treatment of equilibrium for ion pairing. Let鈥檚 derive the fraction of ion pairing for the salt in Box 8-1, which are 0.025FNaCI,Na2SO4,MgCI2,MgSO4. Each case is somewhat different. All of the solutions will be near neutral pH because hydrolysis reactions of Mg2+,SO2-4,Na+,CI-have small equilibrium constants. Therefore, we assume that H+=OH-and omit these species from the calculations. We work MgCI2as an example and then you asked to work each of the others. The ion-pair equilibrium constant, Kipcomes from Appendix J.

Pertinent reaction:

Mg2+CI-MgCI+aqKip=MgCI+aqMgCI+Mg2+Mg2+CI-CI-logKip=0.6.pKip=-0.6A

Charge balance (omitting H+,OH-whose concentrations are both small in comparison with Mg+,MgCI+,CI-:

role="math" localid="1655088043259" 2Mg2-+MgCI+=CI-B

Mass balance:

Mg2-+MgCI+=F=0.025MCCI-+MgCI+=2F=0.050MD

Only two of the three equations (B),(C) and (D) are independent. If you double (c) and subtract (D) , you will produce (B). we choose (C) and (D) as independent equations.

Equilibrium constant expression : Equation (A)

Count : 3 equations (A,C,D) and 3 unknowns Mg2+,MgCI+,CI-

Solve: We will use Solver to find

numberofunknowns-numberofequiliberia=3-1=2unknown concentrations.

The spreadsheet shows the work. Formal concentration F=0.0025Mappears in cell G2. We estimate pMg2+,pCI-in cell B8and B9. The ionic strength in cell B5is given by the formula in cell H24. Excel must be set to allow for circular definitions as described on page role="math" localid="1655088766279" 179. The sizes of role="math" localid="1655088853561" Mg2+,CI-are from Table 8-1and the size of MgCI+is a guess. Activity coefficient are computed in columns E,F. Mass balance b1=F-Mg2+-MGCI+,b2=2F-CI--MgCI+appears in cell H14,H15, and the sum of squares b21+b22 appears in cell H16. The charge balance is not used because it is not independentof the two mass balances.

Solver is invoked to minimizes b21+b22in cell H16be varying pMg2+,pCI-in cells B8and B9. From the optimized concentration, the ion-pair fraction =MgCI+F=0.0815is computed in cell D15.

The problem: Create a spreadsheet like the one for MgCI+to find the concentration, ionic strength, and ion pair fraction in 0.025MNaCI. The ion pair formation constant from Appendix J is log Kip=10-0.5for the reaction Na++CI-NaCIaq. The two mass balances are Na++NaCIaq=F,Na+=CI-Estimate pNa+,pCI- for input and then minimizes the sum of square of the two mass balances.

Using activities, find [Ag+]in 0.060MKSCNsaturated withAgSCN(s)

Find the activity coefficient of each ion at the indicated ionic strength:

(a)SO42-(=0.01M)(b)Sc3+(=0.005M)(c)Ec3+(=0.1M)(d)(CH3CH2)3NH+(=0.05M)

Calculate the ionic strength of (a) 0.008 7 M KOH and (b) 0.000 2 M-La(IO3)3(assuming complete disassociation at this low concentration and no hydrolysis reaction to makeLaOH2+ ).

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