Consider precipitation of Xx-with role="math" localid="1655098271650" Mm+:

xMm++mXx-MxXm(s)Ksp=[Mm+]x[Xx-]mWrite mass balance equations for M and X and derive the equation

VM=Vx0(xCX0+m[Mm+]-x[Xx-](mCM0-m[Mm+]+x[Xx-]where[Xx-]=(Ksp/[Mm+]x)1m

Short Answer

Expert verified

The mass balance for M and X are.

mCMoVM-m[Mm+](VM+Vxo)=xCxoVxo-x[X(x-)](VM+Vxo)

Step by step solution

01

Step 1: Define the mass balance.

Mass balance is the sum of moles of ion and moles of precipitate. It is literally mole balance.

02

Step 2: derive the equation for mass balance M and X.

The equation of mass balance of M and X

VM=VxoCx°+[M+]-[X-]CM°-[M+]-[X-]

The titration reaction of Mm+with role="math" localid="1655099209443" Xx-

xMm++mXx-MxXm(s)mCMo,VMxCx°,Vxo

mCM0is initial concentration of Mm+

VMis volume of addedMm+

xCx0is concentration of Xx-

Vx0is volume ofXx-

The mass balance of M and x .

mCMoVM=m[Mm+](VM+.Vx)+molMXXm(s)mCMoVM=mMm+(VM+Vx)+molMXXm(s)

On rearranging the above equation

we get mCMoVM-m[Mm+](VM+Vxo)=xCxoVxo-x[Xx-](VM+Vxo)

Thus the solution is VM=VxoCx°+[M+]-[X-]CM°-[M+]-[X-]

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Most popular questions from this chapter

A 50.0-mLsample of 50.0-mLis titrated with role="math" localid="1654840699323" 0.0400MCu+.The solubility product of is4.8×10-15. At each of the following volumes of titrant, calculate role="math" localid="1654840898095" pCu+,and construct a graph of pCu+versus milliliters of Cu+added:

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