A) If 25.00 mL of a 0.3000 M H 2 SO 4 solution requires 37.65 mL of KOH to titra

Question

A) If 25.00 mL of a 0.3000 M H2SO4 solution requires 37.65 mL of KOH to titrate, determine the molarity of the KOH ?  (enter answer as regular or scientific notation (e.g. 2.00e-3); do not enter units just numerical answer. Round answer to correct number of significant digits

H2SO4    + 2 KOH K2SO4   + 2 H2O

B) If it takes 30.60 mL of a 0.2000 M NaOH solution to titrate 7.462 grams of vinegar, determine the mmol (millimoles) of acetic acid in the sample? Round off to correct number of significant digits; You may enter as regular or scientific notation (e.g. 2.00e-4) but do not enter units

C) If it takes 30.60 mL of a 0.2000 M NaOH solution to titrate 7.462 grams of vinegar, determine the grams of acetic acid in the sample? Round off to correct number of significant digits; You may enter as regular or scientific notation (e.g. 2.00e-4) but do not enter units

Explanation / Answer

A)

H2SO4    + 2 KOH K2SO4   + 2 H2O

moles of H2SO4 = 25 x 0.3 / 1000 = 7.5 x 10^-3

1 mol H2SO4 ------------> 2 mol KOH

7.5 x 10^-3 mol -----------> ??

moles of KOH = 0.015 mol

moles = molarity x volume

0.015 = Molarity x 37.65 x 10^-3

Molarity of KOH = 0.398 M

B)

mmoles of NaOH = 30.60 x 0.2 = 6.12

mmoles of acetic acid = 6.120 mmol

c)

moles of acetic acid = 6.12 x 10^-3

moles = mass / molar mass

6.12 x 10^-3 = mass / 60.05

mass of acetic acid = 0.3675 g

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