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QJEEAdv06Ch4dfb: JEE Advanced 2006 Chemistry - Class 12 JEEAdv06Ch4dfb Solution: We have taken a saturated solution of AgBr. K sp of AgBr is 12…

We have taken a saturated solution of AgBr. Ksp of AgBr is 12 ×\times 10-14. If 10-7 mole of AgNO3 are added to 1 litre of this solution find conductivity (specific conductance) of this solution in terms of 10-7 S m-1 units. Given, molar conductance of Ag+, Br- and NO3−NO_3^- are 6 ×\times 10-3 Sm2 mol-1, 8 ×\times 10-3 Sm2 mol-1 and 7 ×\times 10-3 Sm2 mol-1

Step-by-Step Verified Solution

AgBr(s)AgBr(s) ⇌\rightleftharpoons Ag+(aq)+Br−(aq)A{g^ + }(aq) + B{r^ - }(aq) ..... [1]

AgNO3(aq)→Ag+(aq)+NO3−(aq)AgN{O_3}(aq) \to A{g^ + }(aq) + NO_3^ - (aq) ...... [2]

Suppose the solubility of AgBr in 10 −- 7 M AgNO3 is s mol L −- 1. Substituting in equation (1) and (2), we get,

∴\therefore Total [Ag+]=(s+10−7)M[A{g^ + }] = (s + {10^{ - 7}})M , Ksp(AgBr)=[Ag+][Br−]{K_{sp}}(AgBr) = [A{g^ + }][B{r^ - }]

or, 12×10−14=(s+10−7)s12 \times {10^{ - 14}} = (s + {10^{ - 7}})s or, s2+10−7s=12×10−14{s^2} + {10^{ - 7}}s = 12 \times {10^{ - 14}}

or, s2+10−7s−12×10−14=0{s^2} + {10^{ - 7}}s - 12 \times {10^{ - 14}} = 0 or, s=3×10−7Ms = 3 \times {10^{ - 7}}M

∴\therefore [Br−]=3×10−7M=3×10−7×103m3=3×10−4m3[B{r^ - }] = 3 \times {10^{ - 7}}M = 3 \times {10^{ - 7}} \times {10^3}{m^3} = 3 \times {10^{ - 4}}{m^3}

[Ag+]=3×10−7+10−7=4×10−7M[A{g^ + }] = 3 \times {10^{ - 7}} + {10^{ - 7}} = 4 \times {10^{ - 7}}M

=4×10−7×103m3=4×10−4m3= 4 \times {10^{ - 7}} \times {10^3}{m^3} = 4 \times {10^{ - 4}}{m^3}

[NO3−]=10−7M=10−7×103m3=1×10−4m3[NO_3^ - ] = {10^{ - 7}}M = {10^{ - 7}} \times {10^3}{m^3} = 1 \times {10^{ - 4}}{m^3}

We know, λ=κC\lambda = {\kappa \over C} or, κ=λ×C\kappa = \lambda \times C .

∴\therefore κBr−=3×10−4×8×10−3S . m−1=24×10−7S m−1{\kappa _{B{r^ - }}} = 3 \times {10^{ - 4}} \times 8 \times {10^{ - 3}}S\,.\,{m^{ - 1}} = 24 \times {10^{ - 7}}S\,{m^{ - 1}}

∴\therefore κAg+=4×10−4×6×10−3S . m−1=24×10−7S m−1{\kappa _{A{g^ + }}} = 4 \times {10^{ - 4}} \times 6 \times {10^{ - 3}}S\,.\,{m^{ - 1}} = 24 \times {10^{ - 7}}S\,{m^{ - 1}}

∴\therefore κNO3−=1×10−4×7×10−3S . m−1=7×10−7S m−1\kappa _{NO_3^ - }^{} = 1 \times {10^{ - 4}} \times 7 \times {10^{ - 3}}S\,.\,{m^{ - 1}} = 7 \times {10^{ - 7}}S\,{m^{ - 1}}

∴\therefore κtotal=κBr−+κAg++κNO3−{\kappa _{total}} = {\kappa _{B{r^ - }}} + {\kappa _{A{g^ + }}} + \kappa _{NO_3^ - }^{}

=24×10−7×24×10−7+7×10−7=55×10−7S m−1= 24 \times {10^{ - 7}} \times 24 \times {10^{ - 7}} + 7 \times {10^{ - 7}} = 55 \times {10^{ - 7}}S\,{m^{ - 1}}

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