Asked by Andrew
A sample of KHC2O4 weighing 0.5839 g is dissolved in 25 mL of water and an endpoint is reached when 31.99 mL of KOH has been added. Calculate the molarity of the KOH solution.
KHC2O4(aq) + KOH(aq) ¨ K2C2O4(aq) + H2O(l)
because the coefficents are all the same, i just assumed that KOH also had 0.5839 g, converted into mols, and divdied by the total volume. however, this answer was wrong...
KHC2O4(aq) + KOH(aq) ¨ K2C2O4(aq) + H2O(l)
because the coefficents are all the same, i just assumed that KOH also had 0.5839 g, converted into mols, and divdied by the total volume. however, this answer was wrong...
Answers
Answered by
DrBob222
I'm not surprised the answer was wrong.
Chemical reactions are by mols and not by grams; therefore, 1 mol of each reactant produces 1 mol of each product but a mol is not a gram.
SO, convert 0.5839 g KHC2O4 to mols, convert mols KHC2O4 to mols KOH (that's the 1:1 part), then calculate the molarity of KOH. Post your work if you get stuck. I expect you will have no trouble.
Chemical reactions are by mols and not by grams; therefore, 1 mol of each reactant produces 1 mol of each product but a mol is not a gram.
SO, convert 0.5839 g KHC2O4 to mols, convert mols KHC2O4 to mols KOH (that's the 1:1 part), then calculate the molarity of KOH. Post your work if you get stuck. I expect you will have no trouble.
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