A) If 25.00 mL of a 0.3000 M H 2 SO 4 solution requires 37.65 mL of KOH to titra
ID: 566367 • Letter: A
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