incomplete answer when i posted earlier, as shown bellow. So lets try again , th

Question

incomplete answer when i posted earlier, as shown bellow. So lets try again , thank you for your guys expertise

.*****Molarity of NaOH = number of moles / Volume in Lit

1 M= number of moles of NaOH/ 36.88 x 10^-3 L

Number of moles of NaOH = 36.88 x 10^-3 *************

4. In an experi mass 12 ss= 128.13 g mol is determined tassium hydrogen oxalate. The following equilibrium results: riment similar to this one, the solubility of potassium hydrogen oxalate (KHC204 molar 8.13 g mol-) is determined. A 5.0 10 M aqueous solution of KCl is saturated with po KHC 0 (s)K (a)+HC204 (aq) Following saturation, the solution is filtered. Portions of the filtered saturated solution are titrated with 0.1172M NaOH solution.

Explanation / Answer

1. Number of moles of NaOH needed:

Determination-1: MxV(L) = 0.1172 mol/L x 36.88 mL x (1L / 1000 mL) = 4.322x10-3 mol

Determination-2: MxV(L) = 0.1172 mol/L x 41.81 mL x (1L / 1000 mL) = 4.900x10-3 mol

2. Number of moles of HC2O4-(aq) titrated:

Since 1 mol of NaOH reacts with 1 mol of HC2O4-(aq), moles of NaOH needed is equal to the number of moles of HC2O4-(aq) titrated. Hence

Determination-1:moles of HC2O4-(aq) titrated = moles of NaOH = 4.322x10-3 mol

Determination-2:moles of HC2O4-(aq) titrated = moles of NaOH = 4.900x10-3 mol

3. [HC2O4-(aq)]sat.sln :

Determination-1: Volume of saturated KHC2O4 solution titrated, V = 25.2 mL = 0.0252 L

Hence [HC2O4-(aq)]sat.sln = moles of HC2O4-(aq) / V = 4.322x10-3 mol / 0.0252 L = 0.1715 M

Determination-2: Volume of saturated KHC2O4 solution titrated, V = 29.1 mL = 0.0291 L

Hence [HC2O4-(aq)]sat.sln = moles of HC2O4-(aq) / V = 4.900x10-3 mol / 0.0291 L = 0.1684 M

4. [K+]initial :

Determination-1: [K+]initial = [KCl] = 5.0x10-2 M

Determination-2: [K+]initial = [KCl] = 5.0x10-2 M

5. [K+] (from dissolved KHC2O4): This amount is equal to the concentration of HC2O4-(aq) in the saturated solution. Hence

Determination-1: [K+] (from dissolved KHC2O4) = 0.1715 M

Determination-2: [K+] (from dissolved KHC2O4) = 0.1684 M

6. [K+]sat.soln: This is equal to the sum of initial K+ concentration and concentration of K+ from dissolved HKC2O4.

Determination-1: [K+]sat.soln = 5.0x10-2 M + 0.1715 M = 0.2215 M

Determination-2: [K+]sat.soln = 5.0x10-2 M + 0.1684 M = 0.2184 M

7. Ksp:

Determination-1: Ksp = [K+]sat.soln x [HC2O4-(aq)]sat.sln = 0.2215 M x 0.1715 M = 3.80x10-2

Determination-2: Ksp = [K+]sat.soln x [HC2O4-(aq)]sat.sln = 0.2184 M x 0.1684 M = 3.68x10-2

8. Average Ksp:

= (3.80x10-2 + 3.68x10-2) / 2 = 3.74x10-2   

9. Solubility of KHC2O4: This is equa to the concentration of HC2O4-(aq) in the saturated solution. Hence

Determination-1: 0.1715 M

Determination-2: 0.1684 M

10: Average solubility: = (0.1715 + 0.1684) / 2 = 0.1700 M

11. Solubility(in g/100 mL): = 0.1700 mol x 128.12 g/mol / 1L = 21.78 g / 1000 mL = 2.178 g/100 mL

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