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Electrochemistry

Question
CBSEENCH12011246

In qualitative analysis, the metals of group I can be separted from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl- concentration is 0.10 M. What will be the concentration of Ag+ and Pb2+ be at equilibrium? (Ksp for AgCl = 1.8 x 10-10, Ksp for PbCl2  = 1.7 x 10-5)

  • [Ag+]  = 1.8 x 10-7 M; [Pb2+] = 1.7 x 10-6 M

  • [Ag+]  = 1.8 x 10-11 M; [Pb2+] = 8.5 x 10-5 M

  • [Ag+]  = 1.8 x 10-9 M; [Pb2+] = 1.7 x 10-3 M

  • [Ag+]  = 1.8 x 10-11 M; [Pb2+] = 8.5 x 10-4 M

Solution

C.

[Ag+]  = 1.8 x 10-9 M; [Pb2+] = 1.7 x 10-3 M

straight K subscript sp space for space AgCl space equals left square bracket Ag to the power of plus right square bracket left square bracket Cl to the power of minus right square bracket

therefore space left square bracket Ag to the power of plus right square bracket space equals space fraction numerator 1.8 space straight x space 10 to the power of negative 10 end exponent over denominator 10 to the power of negative 1 end exponent end fraction

space equals space 1.8 space straight x space 10 to the power of negative 9 end exponent space straight M.

straight K subscript sp space for space PbCl subscript 2 space equals space left square bracket Pb to the power of 2 plus end exponent right square bracket left square bracket Cl to the power of minus right square bracket squared

therefore space left square bracket Pb to the power of 2 plus end exponent right square bracket space equals space fraction numerator 1.7 space straight x space 10 to the power of negative 5 end exponent over denominator 10 to the power of negative 1 end exponent space straight x space 10 to the power of negative 1 end exponent end fraction
space
equals space 1.7 space straight x space 10 to the power of negative 3 end exponent space straight M straight K subscript sp space for space AgCl space equals left square bracket Ag to the power of plus right square bracket left square bracket Cl to the power of minus right square bracket

therefore space left square bracket Ag to the power of plus right square bracket space equals space fraction numerator 1.8 space straight x space 10 to the power of negative 10 end exponent over denominator 10 to the power of negative 1 end exponent end fraction

space equals space 1.8 space straight x space 10 to the power of negative 9 end exponent space straight M.

straight K subscript sp space for space PbCl subscript 2 space equals space left square bracket Pb to the power of 2 plus end exponent right square bracket left square bracket Cl to the power of minus right square bracket squared

therefore space left square bracket Pb to the power of 2 plus end exponent right square bracket space equals space fraction numerator 1.7 space straight x space 10 to the power of negative 5 end exponent over denominator 10 to the power of negative 1 end exponent space straight x space 10 to the power of negative 1 end exponent end fraction
space
equals space 1.7 space straight x space 10 to the power of negative 3 end exponent space straight M