Asked by K
If 8.72 grams of oxalic acid, H2C2O4, a diprotic acid, is neutralized when 23.4 mL of a solution of KOH is added, what is the molarity of KOH?
the topic is acid-base triation, and i have already converted the mass to moles,but I am having trouble finding the volume for the acid (H2C2O4), I am guessing it's just 8.72 mL? Because grams=mL, right? to calculate the molarity i would just divide the moles by volume(in Liters)
the topic is acid-base triation, and i have already converted the mass to moles,but I am having trouble finding the volume for the acid (H2C2O4), I am guessing it's just 8.72 mL? Because grams=mL, right? to calculate the molarity i would just divide the moles by volume(in Liters)
Answers
Answered by
DrBob222
I think that's no and yes but here is what you do. All of these are done stepwise and most follow the same pattern.
1. Write the equation.
H2C2O4 + 2KOH ==> K2C2O4 + 2H2O
2. Convert g H2C2O4 to mols. mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert mols H2C2O4 to mols KOH. That's the value in #2 x 2.
4. Now M KOH = mols KOH/L KOH
You might do well to memorize this four-step process. It will work many problems for you.
1. Write the equation.
H2C2O4 + 2KOH ==> K2C2O4 + 2H2O
2. Convert g H2C2O4 to mols. mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert mols H2C2O4 to mols KOH. That's the value in #2 x 2.
4. Now M KOH = mols KOH/L KOH
You might do well to memorize this four-step process. It will work many problems for you.
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