Question

1. You dissolve 5.00 g sodium hydrogen oxalate (NaHC2O4 ) in suffircient water to produce a final solution volume of 100.0 mL. Assuming all of the solid dissolves, calculate the pH of this solution.

2. You dissolve 5.00 g of sodium oxalate (Na2C2O4) in sufficient water to produce a final solution volume of 100.0 mL. Assuming all of the solid dissolves, calculate the solution pH of this solution.

3. You dissolve 5.00 g of sodium hydrogen oxalate and 5.00 g sodium oxalate in sufficient water to produce a final solution volume of 250.0 mL. Assuming both solids dissolve completely, calculate the pH of this solution.

4. Calculate the pH of the solution prepared in question #3 after 2.00 mL 1.00 M sodium hydroxide is added to it.

5. You dissolve 5.00 g sodium hydrogen oxalate and 1.00 g sodium hydroxide in sufficient water to produce a final solution volume of 250.0 mL. Assuming both solids dissolve completely, calculate the pH of this solution.

6. Calculate the pH of the solution prepared in question #5 after 2.00 mL 1.00 M hydrochloric acid is added to it.

Answer 1

1-

Here the given salt NaHC2O4 is the salt of HC2O4-, which is the conjugate base of the weak acid H2C2O4 (Oxalic Acid).

This salt dissociates as-

NaHC2O4 -----------------> HC2O4- + Na+  

Now since HC2O4- is basic in nature, it can accept H+ from water to form

HC2O4- + H2O ------------> H2C2O4 + OH-  

Now we know Ka for the weak acid (H2C2O4) is 5.37 x 10-2  

So since HC2O4- is the cojugate base of H2C2O4 , we can calculate its dissociation constant (Kb) as-

Kb = Kw / Ka

= 1 * 10-14 / 5.37 x 10-2  

= 0.186 x 10-12  

= 1.86 x 10-13  

Again for the given base, the dissociation constant is calculated by another formula-

Kb = [OH-] * [H2C2O4] / [HC2O4-]  

where all the concentrations are at the time of equilibirum. Now to find these concentrations, we have to from the ICE table-

Given

Mass of NaHC2O4 taken = 5 g

So mols of NaHC2O4 taken = mass / molar mass of NaHC2O4

= 5 g / 112 g/mol

= 0.045 mols

Again total volume of solution = 100.0 mL

= 0.100 L

So initial concentration of NaHC2O4 taken = moles / volume

= 0.045 mols / 0.100 L

= 0.45‬ M

So we can say initial [HC2O4-] = [NaHC2O4] = 0.45‬ M

Reaction	HC2O4- + H2O ------------>	H2C2O4 +	OH-
Initial	0.45‬ M	0	0
Change	-x	+x	+x
Equilibrium	0.45‬ -x	x	x

Now putting the values-

Kb = [OH-] * [H2C2O4] / [HC2O4-]  

(1.86 x 10-13) = [x] * [x] / [0.45‬ -x]  

(1.86 x 10-13) * [0.45‬ -x] = x2  

(0.837 x 10-13) - (1.86 x 10-13)x = x2  

x2 + (1.86 x 10-13)x - (0.837 x 10-13) = 0

Solving this-

x = 2.89 * 10-7  

So we can say, at equilibrium,

[OH-] = x = 2.89 * 10-7 M

So

pOH = -log [OH-] = -log [2.89 * 10-7] = 6.54

So

pH = 14 - pOH = 14 - 6.54 = 7.46