How many moles of AgBr (K_sp = 5 × 10^–13 mol² L^–2) will dissolve in a 0·01 M NaBr solution (NaBr is completely dissociated into Na⁺ and Br^-ion).

Solution

The complete ionisation of NaBr is represented as

NaBr space space space space space space space rightwards arrow space space space space Na to the power of plus space space space space space space plus space space space space Br to the power of minus 0.01 space straight M space space space space space space space space space space space 0.01 space straight M space space space space space space space space space space 0.01 space straight M therefore space space space open square brackets Br to the power of minus close square brackets space equals space 0.01 space straight M space space

The solubility equilibrium of AgBr is represented as

AgBr space space rightwards harpoon over leftwards harpoon space space space Ag to the power of plus space plus space Br to the power of minus straight K subscript sp space plus space open square brackets Ag to the power of plus close square brackets space open square brackets Br to the power of minus close square brackets

Now the product of

space open square brackets Ag to the power of plus close square brackets space open square brackets Br to the power of minus close square brackets

can not exceed 5 x 10^-13 mol² L^-2

open square brackets Ag to the power of plus close square brackets space equals space fraction numerator 5 space cross times space 10 to the power of negative 13 end exponent over denominator 0.01 end fraction space space space space space space space space equals space fraction numerator 5 cross times 10 to the power of negative 13 end exponent over denominator 0.01 end fraction cross times 5 space cross times space 10 to the power of negative 11 end exponent space mol space straight L to the power of negative 1 end exponent

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