Consider the equilibrium in the reaction below, the activities and the Debye-Hucke equation (if necessary), determine...

Consider the equilibrium in the reaction below, the activities and the Debye-Hucke equation (if necessary), determine the solubility of HgBr2 as the concentration of Hg^2+ dissolved in the presence of 0.070 M NaBr. Kps= 1.3 x1-^-9 (HgBr2)

HgBr2(s) <----> HB^+2 + 2Br-

Expert Solution

NaBr is strong electrolyte and it dissociate completely into Na+ and Br-

So,

initially, [Br-] = 0.070 M

HgBr2(s) <----> HB^+2 (aq) + 2Br- (aq)

0 0.070 (initial)

+s +2s (change)

s 0.070+2s (equilibrium)

Ksp = [HB^+2] [Br-]^2

1.3*10^-9 = s* (0.070+2s)

since ksp is small, s will be small and can be ignored as compared to 0.070.

Above expression thus becomes,

1.3*10^-9 = s* (0.070)

s = 1.9*10^-8 M

Answer: 1.9*10^-8 M

queen_honey_blossom answered 2 years ago

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