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Chapter 8: Q20P (page 184)

Write the charge balance for an aqueous solution of arsenic acid, H3AsO4, in which the acid can disassociate to role="math" localid="1654936423245" H3AsO-4,HAsO42-,andAsO43-. Look up the structure of arsenic acid in Appendix G and write the structure ofHAsO2-4

Short Answer

Expert verified

Thus the charge balance equation for aqueous solution of arsenic acid, H3AsO4, in which the acid can disassociate to role="math" localid="1654936784437" H2AsO-4,HAsO42-,andAsO43- isH+=H2AsO4-+2HAsO42-+3AsO43-+OH-

Step by step solution

01

Step 1:Disassociation of an aqueous solution of Arsenic acid.

In this problem, we are in a need to write the charge balance for an aqueous solution of arsenic acidH3AsO4.

We also need to note that the aqueous solution of arsenic acid can disassociate as follows,

i.H3AsO4⇔H++H2AsO4-ii.H3AsO4-⇔H++HAsO42-iii.HAsO42-⇔H++AsO43-iv.H2O⇔H++OH-

02

Step 2:Charge Balance Equation of Arsenic Acid.

The charge balance of the arsenic acid can be written as follows

H+=H2AsO4-+2HAsO42-+3AsO43-+OH-

Thus the charge balance equation for aqueous solution of arsenic acid,H3AsO4, in which the acid can disassociate to H2AsO-4,HAsO42-,andAsO43-is H+=H2AsO4-+2HAsO42-+3AsO43-+OH-

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Most popular questions from this chapter

Find [Hg22+] in equilibrium with 0.010MKCl saturated withHg2Cl2.

Systematic treatment of equilibrium for ion pairing. Let’s 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+aqγMgCI+Mg2+γMg2+CI-γCI-logKip=0.6.pKip=-0.6→A

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=0.025M→CCI-+MgCI+=2F=0.050M→D

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.

Write the charge and mass balances for dissolving CaF2 in water if the reactions are

role="math" localid="1654770961556" CaF2(s)ٟCa2++2FCa2++H2OٟCaOH++H+Ca2+FؚCaF+CaF2(s)ؚCaF2(aq)F+H+ؚHF(aq)HF(aq)+FؚHF2

Why do activity coefficients not appear in the charge andmass balance equations?

Ion pairing. As in Problem 8-30, find the concentration, ionic strength, and ion pair fraction in localid="1654942556135" 0.025FNa2SO4. Assume that the size of theNaSO-4=500pm
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