A. At a certain temperature, the Kp for the decomposition of H2S is 0.722. H2S(g

Question

A. At a certain temperature, the Kp for the decomposition of H2S is 0.722.

H2S(g) = H2(g)+S(g)

Initially, only H2S is present at a pressure of 0.219 atm in a closed container. What is the total pressure in the container at equilibrium?

B. Find the concentration of Hg22 in 0.10 M KI saturated with Hg2I2. Include activity coefficients in your solubility-product expression. The Ksp of Hg2I2 is 4.6× 10–29.

C. Which of the following will be more soluble in an acidic solution than in pure water?

SrSO4

Be(OH)2

AlPO4

PbCl2

CsClO4

Explanation / Answer

A)

H2S = H2+S

Kp = 0.722

PH2S = 0.219

PH2 = 0

PS = 0

in equilbrium

PH2S = 0.219-x

PH2 = 0 + x

PS = 0 + x

Kp = (H2)(S)/(H2S)

0.722 = (x*x)/( 0.219-x)

0.722* (0.219-x) = x^2

x^2 +0.722x -0.158118 = 0

x = 0.1761

for partial pressures:

PH2S = 0.219-x = 0.219-0.1761 = 0.0429

PH2 = 0 + x = 0.1761

PS = 0 + x = 0.1761

Total P = PH2S +PH2 +PS = 0.0429+0.1761+0.1761 = 0.3951 atm

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