Question

A 12.51 mL sample of a saturated solution of Mn(OH)₃ at 17.0 C gave a pH of 9. The same volume of the saturated solution at 80.0 C had a pH 8.465.

1. Calculate the Ksp values of Mn(OH)_³ at the two measured temperatures.

2. Calculate Delta G values for the reaction above. Use gas constant R= 8.31 J/K mol

3. Calculate Delta H & Delta to corresponding to your Delta G calculation(s).

Answer 1

A 12.51 mL sample of a saturated solution of Mn(OH)3 at 17.0 C gave a pH of 9. The same volume of the saturated solution at 80.0 C had a pH 8.465.

Q1

Ksp = [Mn3+][OH-]^3

if [OH-] = S

[Mn3+] = 1/3*[OH-] = 1/3S

Ksp =  [Mn3+][OH-]^3 = (1/3S)(S)^3 = 1/3*S^4

get S

pH = 9, pOH = 14-9 = 5

[OH-] = 10^-pOH = 10^-5

Ksp, @ 9 =  1/3*(10^-5)^4 = 3.33333*10^-21

pH = 8.465, pOH = 14-8.465 = 5.535

[OH-] = 10^-pOH = 10^-5.535 = 2.917*10^-6

Ksp, @ 8.465 =  1/3*(2.917*10^-6)^4 = 2.413*10^-23

Q2

pH of 9.

dG = -RT*ln(K)

dG =-(8.314*298)*ln( 3.33333*10^-21) = 116818.30 ]J/mol = 116.818 kJ/mol

then, pH = 8.465

dG = -RT*ln(K)

dG =-(8.314*298)*ln( 2.413*10^-23) = 129028.447 J/mol = 129.028 kJ/mol

Q3

dG = dH - T*dS

dG = dH - T*dS

then

116.818 = dH - (17+273) * dS

129.028 = dH - (80+273)*dS

dH = 116.818 + (17+273) * dS

substitute in:

129.028 = dH - (80+273)*dS

129.028 = 116.818 + (17+273) * dS- (80+273)*dS

129.028 -116.818 = ((17+273) - (80+273))*dS

12.21 / -63 = dS

-0.19380 = dS

dS = 193.8 J/K

now..

dH = 116.818 + (17+273) * dS

dH =  116.818 + (17+273) * -0.19380

dH = 60.616 kJ/mol