KMnO_4 titrant is used to titrate 0.2400 g of pure sodium oxalate, Na_2C_2O_4. a

Question

KMnO_4 titrant is used to titrate 0.2400 g of pure sodium oxalate, Na_2C_2O_4. according to the reaction: 2Mn()_4 + 5C_2O_4^2 + 16H+ rightarrow 2Mn^2+ + 10CO_2 + 8h_20 if .30.00 ml. of KMn()_4 is required to reach equivalence, what is the molarity of the titrant? (Molar Mass. M: Na_2C_2O_4 = 134.0 g-mol^-1 0.0239 0.0597 0.0746 0.1493 You have 100.0 mL of 0.100 M Na_3PO_4 to which you add 50.0 mL of 0.0950 M NaHiPO_4. What is the pi I of this solution, assuming additive volumes? (Dissociation constants for Hj_3PO_4: pK_1 = 2.15, pK_2 = 7.20, pK_3 = 12.15) 4.10 7.20 11.89 12.19

Explanation / Answer

Molecular weight of Na2C2O4= 134

Moles of Na2C2O4= 0.24/134=0.001791

From the reaction, 5 moles of Na2C2O4 requires 2 moles of KMnO4.

0.001791 moles requires 0.001791*2/5 =0.000716 moles

Molarity= moles/ Volume (L)= 0.000716*1000/30 =.0238M

2. TThe reaction is Na3PO4 + NaH2PO4 ----> 2 Na2HPO4

moles of Na3PO4= 0.1*100/1000 =0.01 and moles of NaH2PO4= 0.095*50/1000=0.00475

limting reactant is Na3PO4. Moles of Na3PO4remaining = 0.01-0.00475=0.00525

pH = pKa + log [Na3PO4]/[Na2HPO4)= 12.15+log (0.00475/0.00525)=12.10 ( close answer is 12.19)

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