Question

In an experiment similar to this one, the solubility of potassium hydrogen oxalate (KHC2O4, molar mass= 128.13 g mol) is determined. A 5.0 10^-2 M aqueous solution of KCI is saturated with potassium hydrogen oxalate. The following equilibrium results: KHC2O4(s)-----K^+ + HC2O4^-

Following saturation, the solution is filtered. Portions of the filtered saturated solution are titrated with 0.1172M NaOH solution.

The following date are obtained:

Volume of saturated KHC2O4 Solution Titrated, mL Determination #1 25.2 Determination #2 29.1

Volume of NaOH solution used, mL ---------------------Determination #1 36.88 Determination #2 41.81

Temperature of saturated solution, C-------------------- Determination #1 20.1 Determination #2 20.1

Calculate the following:

number of moles off NaOH needed, mol _______________________For both determinations

number of moles of HC2O4 ^- titrated, mol______________________For both determinations

[HC2O4^-] sat. sol'n_______________________________________For both determinations

[K^+]initial______________________________________________For both determinations

[K^+]from dissolved KHC2O4________________________________For both determinations

[K^+] sat. sol'n___________________________________________For both determinations

Ksp___________________________________________________For both determinations

average Ksp

solubility of KHC2O3 in 5.0 x 10^-2M KCI, mol L___________________for both determinations

Average solubility of KHC2O4 in 5.0 x 10^-2 KCI, mol L

solubility of KHC2O4 in 5.0 x 10^-2 M KCI, g per 100 mL

Answer 1

Numer of moles of NaOH needed:

detemination#1, 36.88ml: NaOH needed=

                                               number of mole= 0.1729/.40= 0.0043 mole

determination#2, 41.81ml: NaOH needed

                                               number of mole= 0.196/40=0.0049mole

Number of moles of HC₂O₄^- titrated:

First of all the strength of oxalate need to be calculated from the two NaOH concentrations using V₁S₁=V₂S₂ formula.

S_oxalate= 0.1172X36.88/25.2=0.1715M and S_oxalate= 0.1172X41,81/29,1=0.1684M

determination#1, 25.2 ml: oxalate titrated=

                        number of mole= 0.554/128.13= 0.0043 mole

determination#2, 29.2 ml:

                          number of mole= 0.627/128.13= 0.0049 mole

Conentration of K⁺ from dissolved potassium oxalate is equal to the concentration of potassium oxalate

so for determination#1: concentration of K⁺ = 0.1715 M

determination#2: concentration of K⁺ = 0.1684 M

Concentration of K⁺ in saturated solution is the summation of K⁺ from KCl and from that of potassium oxalate

for determination#1: total K⁺ in saturated solution= 0.05 M+0.1715 M=0.2215 M

determination#2: total K⁺ in saturated solution= 0.05 M+0.1684 M=0.2184 M