A. mols succinic acid in the initial 1.700 g is mols = grams/molar mass = ?
Then you took 1/10 of that with
?mols x (25.00 mL/250.00 mL) = ?
B. If we call succinic acid H2C, then
H2C + 2NaOH ==> Na2C + 2H2O
mols succinic acid from a = ?
2x that = mols NaOH
M NaOH = mols NaOH/L NaOH. Solve for L NaOH and convert to mL
C. See B above.
A student weights 1.700 g of succinic acid and dissolves it in water in a 250.0 mL volumetric flask. A 25.00 mL sample of this solution is withdrawn and placed in a 125 mL Erlenmeyer flask, to which 5 drops of the acid-base indicator phenolphthalein is added. This solution is titrated with a sodium hydroxide solution of unknown molarity in a buret.
a) How many moles of succinic acid were placed in the Erlenmeyer flask?
b) If the unknown sodium hydroxide solution was known to be approximately 0.20 M, at approximately what buret volume would you expect the end point of the titration?
c) If the measured volume at the end point was 13.32 mL of sodium hydroxide added, calculate the actual molarity of the sodium hydroxide solution.
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