Question

Please help answer questions 1-4 of my assignment with the given [HCl] Concentration of 0.1522

Titration of Room Temperature Ca(OH)2 Solution. Note these are 10.00 mL samples. Room Temperature_210 "C294 K (lab manual, p. 8-5) Trial Trial 2 Trial 3 Final Volume (mL) Initial Volume (mL) Volume Delivered (mL) Average Volume (mL) 28.66 mL Titration of "Hot" Ca(OH)2 Solution. Note these are 10.00 mL samples. Temperature of "Hot" Solution_90 _°C - 363 _K (lab manual, p. 8-5) Trial 1 Trial 2 Trial 3 Final Volume (mL) Initial Volume (mL) Volume Delivered (mL) Average Volume (mL) 18.05 mL The remainder of the data analysis will use the data you have acquired to determine the specified thermodynamic quantities (See pp. 8-5 & 8-6 for additional background information.) 1. Start by using the stoichiometry of the titration reaction to determine the concentration of OH ions in each of the saturated Ca(OH)2 solution samples that were titrated above. The balanced titration reaction is simply the neutralization reaction: H30+ (aq) + OH(aq) + 2 H200). The key piece of information here is the one-to-one ratio of hydronium to hydroxide. Remember to use the diluted HCl concentration from the bottom of p. 8-8, and that you need to find the molar concentration of OH in the 10.00 mL samples of the saturated Ca(OH)2 solutions. Room Temperature [OH-] mol L' High Temperature [OH-] mol L-

Answer 1

The reactions involved are:
$Ca(OH)_{2}(s)\, \rightleftharpoons Ca^{+2}(aq)\, +\, 2OH^{-}(aq)$
Let solubility of Ca(OH)2 = S
At equilubrium,
[Ca+2] = S
[OH-] = 2S
This base is titrated against HCl.
$Ca(OH)_{2}(aq)\, +\, 2HCl(aq)\, \rightarrow \, CaCl_{2}(aq)\,\, +\, H_{2}O(l)$
Equivalence point calculations at 294 K:
(m.Eq) of Ca(OH)2 = (m.Eq) of HCl
(M x v x n-factor) of Ca(OH)2 = (M x v x n-factor) of HCl
(M x 10.00 mL x 2) of Ca(OH)2 = (0.1522 M x 28.66 mL x 1) of HCl
M of Ca(OH)2 = (0.1522 M x 28.66 mL x 1)/(10.00 mL x 2) = 0.218 M
Hence, S of Ca(OH)2 = 0.218 M
A saturated solution represents a state of equilibrium.
Keq = Ksp
Ksp = [Ca+2][OH-]2= S(2S)2 = 4S3 = 4 x ( 0.218 )3 = 0.04144
NOTE: Using K = 1/2[OH-]3 as mentioned in the point 2 of the procedure will also give the same results.
K = 1/2(2S)3 = 1/2 x 8S3 = 4S3
$\Delta G_{c}\, =\, -\, 2.303\, R\, T\, log K$
$\Delta G_{c}\, =\, -\, 2.303\, \times \, 8.314\, \times \, 294\, \times \, log \, 0.04144\, =\,7.779\, \times \, 10^{3}\, J$
Equivalence point calculations at 363 K:
(m.Eq) of Ca(OH)2 = (m.Eq) of HCl
(M x v x n-factor) of Ca(OH)2 = (M x v x n-factor) of HCl
(M x 10.00 mL x 2) of Ca(OH)2 = (0.1522 M x 18.05 mL x 1) of HCl
M of Ca(OH)2 = (0.1522 M x 18.05 mL x 1)/(10.00 mL x 2) = 0.1373 M
Hence, S of Ca(OH)2 = 0.1373 M
A saturated solution represents a state of equilibrium.
Keq = Ksp
Ksp = [Ca+2][OH-]2= S(2S)2 = 4S3 = 4 x ( 0.1373 )3 = 0.01035
$\Delta G_{h}\, =\, -\, 2.303\, R\, T\, log K$
$\Delta G_{h}\, =\, -\, 2.303\, \times \, 8.314\, \times \, 363\, \times \, log \, 0.01035\, =\,1.3796\, \times \, 10^{4}\, J$
$\Delta G\, =\,\Delta H\, -\, T \Delta \, S$
The values of $\Delta$ H and $\Delta$ S remain unchanged over aconsiderable range of temperature.  
$\Delta G_{c}\, =\,\Delta H\, -\, T_{c} \Delta \, S\: \: \: \: \: Eq-(1))$
$\Delta G_{h}\, =\,\Delta H\, -\, T_{h} \Delta \, S\: \: \: \: \: Eq-(2))$
$7.779\, \times \, 10^{3}\, =\,\Delta H\, -\, 294 \Delta \, S\: \: \: \: \: Eq-(1))$
$1.3796\, \times \, 10^{4}\, =\,\Delta H\, -\, 363 \Delta \, S\: \: \: \: \: Eq-(2))$
Solving the pair of linear equations:
$\Delta$ S = - 87.202 J/K
$\Delta$ H = -17858.388 J = - 17.858 kJ