Question

Given the diprotic acids information summarized in Table 4, predict the titration volumes for the first and second equivalence points of titrations using a 0.04 M standardized NaOH reagent. Enter your predictions into the corresponding rows of Table 5. Note that these diprotic acids constitute the set of materials from which the acid unknowns were selected. Also note that superscripted numbers are source reference citations.

Acid Name	MW (g/mol)	Solubility (g/100 mL H20)	pka1	pka2
Oxalic (dihydrate)	126.07	14.3	1.23	4.19
o-phthalic	166.14	0.6	2.95	5.41
malonic	104.06	154	2.83	5.69
succinic	118.09	5.8	4.17	5.61
maleic	116.07	47.88	1.83	6.07

Acid Name	Acid (g)	Acid Moles	NaOH Volume at 1st Equiv.	NaOH volume at 2nd Equiv
Oxalic (dihydrate)	0.3245
o-phthalic	0.3045
malonic	0.3238
succinic	0.3103
maleic	0.3105

Answer 1

Formulas: mass Moles = molar mass moles of acid molarity of NaOH 2x moles of acid molarity of NaOH NaOH volume at 1st Equiv. (in mL ) = x1000 NaOH volume at 2nd Equiv.(in mL) 1000 Acid Moles NaOH volume at 1st NaOH volume at 2nd Acid Name Acid (g) Equiv Equiv 0.3245 g 126.07 g/mol 0.3045g 166.14 g/mol 0.3238 g 104.06 g/mol 0.3103 g 118.09 g/mol 0.3105 g 116.07 g/mol Oxalic 0.3245 0.00257 mol 64.25 mL 128.5 mL o-phthalic0.3045 0.00183 mol 45.75 mL 91.5 mL malonic 0.3238 - 0.003 11 mol 77,75 mL 155.5 mL succinic 0.3103 0.00263 mol 65.75 mL 131.5 mL maleic 0.3105 0.00268 mol 67.00 mL 134.0 mI