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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+FƶCaF+CaF2(s)ƶCaF2(aq)F+H+ƶHF(aq)HF(aq)+FƶHF2

Short Answer

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The process's mass balance is determined by CaF2gives 2molFfor each molCa.

role="math" localid="1654772123295" F-+CaF++2CaF2aq+HF-+2HF2-ÁåŸspeciescontainingF=2Ca2++CaOH++CaF++2CaF2aqÁåŸspeciescontainingCa2+

Step by step solution

01

Definition of calcium fluriode CaF2

  • Calcium fluoride (CaF2) is an inorganic compound made up of the elements calcium and fluorine.
  • It's an insoluble white solid. Fluorite (also known as fluorspar) is the mineral form, which is often richly coloured due to impurities.
02

Determine the charge and mass balances for dissolving CaF2

The dissolving charge and mass balancesCaF2It must be given in water.

Balance of charge: [C] -concentration of a cation and n-charge of the cation.

The [A] -concentration of an anion and the m-magnitude of the anion's charge.

The total number of species in a solution containing a certain atom (or group of atoms) must equal the total amount of that atom (or group) given to the solution (mass balance)..

In the given information,

CaF2(s)ٟCa2++2FCa2++H2OٟCaOH++H+Ca2+FƶCaF+CaF2(s)ƶCaF2(aq)F+H+ƶHF(aq)HF(aq)+FƶHF2

On the basis of electroneutrality, charge balance:

role="math" localid="1654771916727" F-+HF2-+OH-=2Ca2++CaOH++CaF++H+

By definition, the process's mass balance is determined by CaF2

gives 2 m ol F for each molCa.

F-+CaF++2CaF2aq+HF-+2HF2-ÁåŸspeciescontainingF=2Ca2++CaOH++CaF++2CaF2aqÁåŸspeciescontainingCa2+

Charge and mass balances for calcium fluoride dissolutionCaF2 it was provided in water.

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Most popular questions from this chapter

Sodium acetate hydrolysis treated by Solver with activity coefficients.

(a) Following the NH3 example in Section 8-5, write the equilibria and charge and mass balances needed to find the composition of 0.01 M sodium acetate (Na+A-). Include activity coefficients where appropriate. The two reactions are hydrolysis (pKb = 9.244) and ionization of H2O.

(b) Including activity coefficients, set up a spreadsheet analogous to Figure 8-12 to find the concentrations of all species. Assign an initial value of ionic strength = 0.01. After the rest of the spreadsheet is set up, change the ionic strength from the numerical value 0.01 to the correct formula for ionic strength. This two-step process of beginning with a numerical value and then going to a formula is necessary because of circular references between ionic strength and concentrations that depend on ionic strength. There are four unknowns and two equilibria, so use Solver to find 4 - 2 = 2 concentrations (pC values). Solver does not find both pC values at the same time well in this problem. Execute one pass to find both pC values by varying pA and pOH to minimizeΣbi2 . Then vary only pA to minimizeΣbi2 . Then vary only pOH to minimize Σbi2. Continue alternating to solve for one value at a time as long as Σbi2 continues to decrease. Find [A-], [OH-], [HA], and [H+]. Find the ionic strength, pH =-log([H+] γ+) and the fraction of hydrolysis = [HA]/F.

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