CBSE Class 11-science Answered

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Asked by Prashant DIGHE | 12 Jun, 2020, 09:53: PM

Expert Answer

Solubility product of AgCl can be calculated by-

space space space space space space space space space space space space space space space space space space space space space space space space space space space space Na subscript 2 CO subscript 3 space plus space space 2 AgCl space rightwards arrow 2 NaCl space plus space Ag subscript 2 CO subscript 3 Mili space mole space added space space space space space space 1.5 cross times 5 space space space space space Excess space space space space space space space space 0 space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 7.5 Milli space mole space left space space space space space space space space space space 7.5 minus straight a space space space space space space space space space space space space space space space space space space space space space space space 2 straight a space space space space space space space space space space space space space space straight a left square bracket Cl to the power of minus right square bracket equals fraction numerator 0.0026 over denominator 35.5 end fraction equals 7.32 cross times 10 to the power of negative 5 end exponent space straight M Concentration space of space Cl to the power of minus space formed equals fraction numerator Milli space mole over denominator Volume space in space mL end fraction equals fraction numerator 2 straight a over denominator 5 end fraction fraction numerator 2 straight a over denominator 5 end fraction equals 7.32 cross times 10 to the power of negative 5 end exponent space straight M straight a equals 1.83 cross times 10 to the power of negative 4 end exponent space space milli minus mole Milli space mole space of space Na subscript 2 CO subscript 3 space left space in space 5 space ml equals 7.5 minus 1.83 cross times 10 to the power of negative 4 end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space almost equal to 7.5 open square brackets CO subscript 3 superscript negative 2 end superscript close square brackets equals fraction numerator 7.5 over denominator 5 end fraction equals 1.5 straight K subscript sp equals open square brackets Ag to the power of plus close square brackets squared space open square brackets CO subscript 3 superscript negative 2 end superscript close square brackets space space 8.2 cross times 10 to the power of negative 12 end exponent space space equals open square brackets Ag to the power of plus close square brackets squared cross times 1.5 open square brackets Ag to the power of plus close square brackets squared equals 5.46 cross times 10 to the power of negative 12 end exponent open square brackets Ag to the power of plus close square brackets equals 2.34 cross times 10 to the power of negative 6 end exponent straight K subscript sp space for space AgCl equals open square brackets Ag to the power of plus close square brackets open square brackets Cl to the power of minus close square brackets space straight K subscript sp equals left parenthesis 2.34 cross times 10 to the power of negative 6 end exponent right parenthesis left parenthesis 7.32 cross times 10 to the power of negative 5 end exponent right parenthesis straight K subscript sp equals 1.71 cross times 10 to the power of negative 10 end exponent space left parenthesis mole space divided by litre right parenthesis to the power of negative 2 end exponent space space space space space space space space space

Answered by Ravi | 13 Jun, 2020, 10:25: PM

CBSE Class 11-science Answered

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CBSE Class 11-science Answered

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