A titration of 5.00 mL acetic acid required 16.17 mL of 0.2075 M sodium hydroxid

Question

A titration of 5.00 mL acetic acid required 16.17 mL of 0.2075 M sodium hydroxide for complete neutralization. Calculate the molarity (moles/D of the acetic acid. Note: since the mole ratio between acetic acid and sodium hydroxide is 1:1, the following equation applies MaVa = MbVb, where M = molarity (moles/L), V = volume, a = acid (acetic acid), b= base (sodium hydroxide). Show the full calculation and use proper units and sig figs. (b)convert the answer in 1a from moles/L to g/L. (c) A titration of a sample of an unknown monoprotic acid required 35.14 mL of 0.2075 M NaOH to reach the endpoint. Calculate the number of moles of NaOH added. Show the full calculation and use proper units and sig figs.

Explanation / Answer

1)

the reaction

CH3COOH + NaOH --> CH3COONa + H20

we can see that

mole ration between CH3COOH and NaoH is 1:1

also

moles = molarity x volume

so

MaVa = MbVb

Ma x 5 = 0.2075 x 16.17

Ma = 0.671055

so

molarity of acetic acid is 0.671055 moles/L

b)

now

concentration in g/L = conc in mol/L x molar mass

molar mass of acetic acid = 60

so

concentration in g/L = 0.671055 x 60

concentration in g/L = 40.2633

c)

we know that

moles = molarity x volume (ml) / 1000

given

molarity of NaoH = 0.2075

volume = 35.14 ml

so

moles of NaOH added = 0.2075 x 35.14 /1000

moles of NaOH added = 7.29155 x 10-3

so

7.29155 x 10-3 moles of NaOH are added

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