Asked by Cassie
A student titrated a solution containing 3.7066 grams of an unknown triprotic acid to the end point using 28.94 milliliters of 0.3021 M KOH. What is the molar mass of the unknown acid? And write a balanced equation for the reaction.
Answers
Answered by
DrBob222
You don't say to which end point the triprotic acid was titrated. It could be titrated to the first H, the second, or all three. I will assume all three H ions were neutralized.
H3A + 3KOH ==> K3A + 3H2O
moles KOH = M x L = ??
Using the coefficients in the balanced equation, convert mole KOH to moles H3A. That will be ??moles KOH x (1 mole H3A/3 moles KOH) = (1/3) x moles KOH.
Then moles H3A = grams H3A/molar mass H3A. YOu know moles and grams H3A, solve for molar mass.
H3A + 3KOH ==> K3A + 3H2O
moles KOH = M x L = ??
Using the coefficients in the balanced equation, convert mole KOH to moles H3A. That will be ??moles KOH x (1 mole H3A/3 moles KOH) = (1/3) x moles KOH.
Then moles H3A = grams H3A/molar mass H3A. YOu know moles and grams H3A, solve for molar mass.
Answered by
vergtyu
1272
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