A solution of malonic acid H2C3H2O4, was standardized by titration with 0.100 M NaOH solution. If 20.76 ml of the NaOH solution were required to neutralize completely 13.15 ml of malonic acid solution, what is the molarity of the malonic acid solution?
2006-10-31 20:58:03 · 3 answers · asked by Catherine D 1 in Science & Mathematics ➔ Chemistry
H2C3H2O4 + 2NaOH --> Na2C3H2O4 + 2H2O
mole of NaOH needed = 0.1 * 20.76 = 2.76 mmole
mole of malonic acid = 1.38 mmole
molarity of malonic acid = 1.38 / 13.15 = 0.104 M
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vixklen
2006-10-31 22:45:35 · answer #1 · answered by vixklen 3 · 1⤊ 0⤋
The balanced equation is:
CH2(COOH)2 + 2NaOH --> CH2(COONa)2 + 2H2O
Find the moles of NaOH:
n = C*V = 0.1*20.76x10^(-3) = 20.76x10^(-4) mol
2 moles of NaOH react with 1 mol of malonic acid, so
20.76x10^(-4) mol of NaOH react with 20.76x10^(-4)/2 = 10.38x10^(-4) mol of malonic acid
Now find the molarity of the malonic acid solution:
C = n/V = 10.38x10^(-4)/13.15x10^(-3) = 0.079 M approx.
2006-10-31 23:51:48 · answer #2 · answered by Dimos F 4 · 1⤊ 0⤋
moles NaOH added = concentration x volume / 1000
moles NaOH = 0.1 x 20.76 /1000 = 0.002076 moles
from equation
1mole NaOH = 1 mole malonic acid
0.002076 moles NaOH = 0.002076 moles malonic acid
concentration malonic acid=moles x 1000 /volume
conc malonic acid = 0.002076 x1000 / 13.15 = 0.158 M
2006-10-31 23:30:42 · answer #3 · answered by john f 1 · 1⤊ 1⤋