What is Ka for H3PO4(aq) ⇌ H+(aq) + H2PO4-(aq) a. Ka = (H3PO4)/(h+)(H2PO4-) B. Ka = (H3PO4)(H+)(H2PO4-) C. Ka = (H3PO4)/(H+)(H2P
1 answer:
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Answer:
M = 0.011
Half reactions for reduction and oxidation:
oxidation:
HC12H6Cl2O2N + NH+ +2e- = HC12H8Cl2O2N
reduction:
C6H8O6 = C6H6O6 + 2H+ + 2e-
Explanation:
Molarity of the ascorbic acid solution:
Ascorbic acid molar mass = 173.12 g/mol
Then you have to find the mols:
mol = 2g/176,12 =0.011 mol Ascorbic acid.
Use the molarity equation:
Molarity (M) = 0.011 mol / 1 L = 0.011 M
Reactions for the reduction of DCPIP by ascorbic acid:
These reactions have to be balanced by the oxide reduction method.
Because it involves the transfer of electrons from ascorbic acid to the DCPIP.
Remember: when is happens a reduction, also happens oxidation.
To see the reactions watch the images attached.
The first one has the calculus for the molarity of the ascorbic acid solution.
The second one has the reactions for the reduction and oxidation of the compounds.
