You are trying to figure out the thermodynamic data for Al(OH)_3 and you test tw

Question

You are trying to figure out the thermodynamic data for Al(OH)_3 and you test two different temperatures to collect data. You find that the pH at 25.0 degree C is 8.261 and at 50.0 degree C the pH reading was at 1.412. Show all calculations and solve for the following values with correct units and significant figures. R = 8.314 times 10^-3 kJ/mol K_sp (25.0 degree C) = _____ K_sp (51.4 degree C) = _____ Delta G (25.0 degree C) = _____ Delta G (51.4 degree C) = _____ Delta S degree = _____ Delta H degree = _____

Explanation / Answer

a)

Ksp = [Ca+2][OH-]^2

[OH-] = 10^-pOH = 10^-(14-8.3261) = 0.00000211884 M

[Ca+2] = 1/2*[OH-] = 1/2*0.00000211884 = 0.00000105942 M

so

T = 25°C

Ksp = (0.00000105942)(0.00000211884^2) = 4.7562*10^-18

b)

Ksp = [Ca+2][OH-]^2

[OH-] = 10^-pOH = 10^-(14-1.412) = 2.582*10^-13 M

[Ca+2] = 1/2*[OH-] = 1/2*2.582*10^-13 = 1.291*10^-13 M

so

T = 50°C

Ksp = ( 1.291*10^-13)*((2.582*10^-13)^2) = 8.606*10^-39

c)

dG = -RT*ln(Ksp)

dG = -8.314*298*ln(4.7562*10^-18)

dG = 98823.11 J/mol

dG = 98.82 kJ/mol

d)

dG = dG° + RT*ln(Ksp)

dG = 98823.11 - 8.314*(50+273) * ln(8.606*10^-39)

dG =334195.94

finally

dG = dH - T*dS

dG =  dH - T*dS

substitute

98823.11 = dH - 298*dS

334195.94 =  dH - 348*dS

solve

(334195.94 -98823.11 ) = (-348 + 298)*dS

dS = (334195.94 -98823.11 ) /(-348 + 298)

dS = -4707.45 J/molK

so

334195.94  =  dH - 348*dS

334195.94  =  dH - 348*(-4707.45 )

dH = 334195.94 + 348*(-4707.45 )

dH = -1303996.66 J/mol

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.