Question

A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its Ksp at 25 °C. The endpoint was reached when 45.00 mL of the manganese(II) hydroxide solution was titrated with 4.88 mL of 0.0050 M HCl solution. What is the Ksp of manganese(II) hydroxide?

expression:(ΔGo = -RT ln K )How does the magnitude of the equilibrium constant relate to the ΔGo? How does the magnitude of the equilibrium constant relate to the deltaG?

a. Smaller positive K = larger negative ΔGo

b. Larger negative K = larger negative ΔGo

c. Larger positive K = larger positive ΔGo

d. Larger positive K = larger negative ΔGo

Answer 1

For this problem first we need to write reaction between manganese hydroxide and hydrochloric acid.

Mn(OH)₂ (s) + 2 HCl (aq) -----> MnCl₂ (aq) + 2 H₂O (l)

Now by using dilution law we need to find concentration of Mn(OH)₂. By using dilution law,

M₁V₁ = M₂V₂

M₁ * 45.00 ml = 0.0050 M * 4.88 ml

M₁ = 0.0050 M * 4.88 ml / 45.00 ml

M₁ = 0.0005422 M Mn(OH)₂

Now write the equilibrium equation for Mn(OH)₂ since it is less soluble in water.

Therefore, Ksp = [Mn²⁺] [OH^-]²

Now [Mn²⁺]= 0.0005422 M and [OH^-] = 2*0.0005422 = 0.0010844 M

Ksp = 0.005422 * (0.0010844)² = 0.005422 * 1.17592336 * 10^-6 = 6.37585 * 10^-9

Now use this Ksp in formula of delta G which is,

delta G = -RT lnK

T = 25^oC = 25+273.15 = 298.15 K

ΔG = -8.314 J/ mol K*298.15 K *l n(6.37585 * 10^-9)

= -8.314 J/mol * 298.15 * (-18.870747)

ΔG = 46777.1680 J/mol

From above calculation, it is seen that the value of K is very less as compared to delta G. Therefore we can say that,

Smaller the positive K = the larger will be ΔG.

If you find any mistake in this please mention in the comment box.

Thanks.