Question

Oxalic acid, H.C.O is a diprotic acid. What is the pH of a 0.100 M solution of oxalic acid, pk. - 1.25 pk = 4.14. Hard water contains a relatively high concentration of multivalent ions. A common ion found in hard water is Capt, and one of the sources is Caso.. What is the Ksp of calcium sulfate if the molar solubility of CaSO, is 0.0030 M? The solubility of CaF2 in pure water at 25°C is 0.0011 M. What is the molar solubility of CaF2 in a solution that also contains 0.150 M NaF? Imagine 15.0 g of ice at 0°C (273 k) is placed in a room at 25°C (298 K). Calculate ASuniverse for the process of ice melting at 25°C; Htus for water is 6.02 kJ/mol. Consider the combustion of propane (C3Hb), and calculate the Gibbs free energy change in kJ/mol at 25°C. CzHa(g) + 502(g) - 300(g) + 4H2O(g) Hon=-2,044 kJ/mol; ASn = 100.3 /(mol)

5 questions please show work

Answer 1

pka1 = 1.25 $\therefore$ -logka1 =pka1=1.25 ka1 =5.625 x 10^-2    pka2 = 4.14  $\therefore$ -logka2 =pka2=4.14 ka2 =1.380 x 10^-5

ka2<<<<ka2   $\therefore$  ka2 can be neglected

weak acids like oxalic acid dissociate to a very  small extent for weak acid [H+] =$\sqrt{}$ka c

[H+] =$\sqrt{}$5.625x10^-2 x 0.1 =7.5 x 10^-2

PH =-log  [H+] = -log 7.5 x 10^-2 = 1.1249

dissociation of CaSO4 can be represented as    CaSO4 $\rightleftharpoons$ Ca+2 + SO4-2

S S where S =solubility

Ksp of   CaSO4 = S2 =.0030^2 =   9x 10^-6

dissociation of   Ca F2 can be represented as Ca F2 $\rightleftharpoons$ Ca+2 + 2 F-1

S 2S

   Ksp of   CaF2 =   [ Ca+2 ] [ F-1]2 = S X 2S2 = 4S3 = 4X( 1.1 X 10^-3)3 = 1.331 X 10^-9

  dissociation of Na F can be represented as Na F $\rightarrow$ Na+ + F-1

0.15M 0 0

0 0.15 0.15

[ F-1] from Na F = 0.15   [ F-1] from Ca F2 Is negligible

Ksp of   CaF2 =   [ Ca+2 ] [ F-1]2 1.331 X 10^-9 = S X 0.152   Solubilityof   CaF2 in Na F =1.331x10^-9/0.152 = 5.9155 x 10^-9  

$\Delta$ Suniverse = $\Delta$ Ssystem +$\Delta$ Ssurroundings $\Delta$ Hfus   for water =6.02 KJ/mole for 15g of ice $\Delta$ Hfus=15 x6.02/18 KJ = 5.01666 KJ =5016.66J

$\Delta$ Suniverse =5016.66/273 + - 5016.66/298 = 18.376-16.834 = 1.542 J $\Delta$ Ssystem =qrev/T

  $\Delta$ G = $\Delta$ H-T$\Delta$S

Gibbs free energy change,$\Delta$G=-2044X1000 - (298 X $\Delta$ S)

= 2.0141 X106 J = 2.0141 X 1O3​​​​​​​ KJ