What is the [S^2-] in a solution that is saturated with hydrogen sulfide and the

Question

What is the [S^2-] in a solution that is saturated with hydrogen sulfide and the pOH = 7.5. Given: K_H2s = 1.0 Times 10^-27 = [H^+]^2 [S] Given a 0.10 M NaCN solution. Write the net ionic equation for the hydrolysis reaction. Given a 0.10 M NaCN solution. Write the mathematical expression for law of chemical equilibrium involving hydrolysis. Consider a 0.10 M solution of NaBrO. What is the numerical value of the hydrolysis constant, K_H? K_HBro = 2.0 Times 10^-9. What is the pH of a .007 M solution of KCN? K_HCN = 4.0 Times 10^-10

Explanation / Answer

Q1.

Strategy:

a) State the Equilibrium between H2S, H+ & S-2

b) Relate pOH to OH- ions, then Apply Kw for H+ and OH- ratio

c) Solve for [S-2] from both expressions.

a)

Relate with the equilibrium:

K = [H+]^2[S-2]

K = 10^-27

so

10^-27 = [H+]^2[S-2]

Clearly, we have 1 equation with 2 unknown. H+ and S-2 concnetrations

We need to relate H+ from pOH

b)

pOH = -log([OH-])

pOH = 7.5 ( given data)

solve for [OH-]

[OH-] = 10^-pOH = 10^-7.5

[OH-] = 3.1622*10^-8 M

Now... Use the constant of equilibrium of water, Kw, for [H+] concentration

Kw = 10^-14 always valid at 25C for pure water

Kw = [H+][OH-]

10^-14 = ([H+])(3.1622*10^-)

solve for [H+]

[H+] = (10^-14)/(3.1622*10^-8) = 3.1623*10^-7 M

Note that we now have 1 eqution with 1 unknown (S-2)

c)

relate both equations

10^-27 = ([H+]^2)([S-2])

10^-27 = ((3.1623*10^-8)^2)([S-2])

[S-2] = (10^-27)/ ((3.1623*10^-8)^2)

[S-2] = 9.9998587*10^-13 M

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