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3 July, 11:49

The solubility of agcl (s) in water at 25 ∘c is 1.33*10-5mol/l and its δh∘ of solution is 65.7 kj/mol. what is the solubility at 47.7 c

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  1. 3 July, 14:45
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    According to this equation:

    AgCl (s) ↔ Ag + (aq) + Cl - (aq)

    so K1 = [Ag+][Cl-]

    when [Ag+] = [Cl-] we can assume both = X

    and when we have X the solubility = 1.33 x 10^-5 mol / L

    by substitution:

    ∴ K1 = X^2

    = (1.33 x 10^-5) ^2

    = 1.77 x 10^-10

    by using vant's Hoff equation:

    ln (K2/K1) = (ΔH/R) * (1/T2-1/T1)

    when ΔH = 65700 J / mol

    R = 8.314

    T1 = 25+273 = 298 K

    T2 = 47.7 + 273 = 320.7

    by substitution:

    ∴㏑ (K2/1.77 x 10^-10) = (65700/8.314) * (1/320.7 - 1 / 298)

    by solving for K2

    ∴K2 = 2.7 x 10^-11

    and when K2 = X^2

    ∴ the solubility X = √ (2.7 x 10^-11)

    = 5.2 x 10^-6 mol/L
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