If a 16.5 mL sample of 1.1 M solution of each of the following acids is reacted with 0.75 M NaOH, how many milliliters of the NaOH are required for the titration? What is the total volume (in mL) of solution at the equivalence point?
16.5 ml H2SO4 titrated with 0.75 M NaOH
Volume of NaOH (ml) 24.2 (incorrect)
Total Volume (ml) 40.7 (incorrect)
H2SO4 + 2NaOH ----------------> Na2SO4 + H2O
1 mole 2 moles
H2SO4 NaOH
M1 = 1.1M M2 = 0.75M
V1 = 16.5ml V2 =
n1 = 1 n2 = 2
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 1.1*16.5*2/0.75*1 = 48.4ml
volume of NaOH = 48.4ml
total volume = 16.5 +48.4 = 64.9ml
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50.0 mL sample of the weak acid
the concentration of the weak acid = 0.15 M
25 mL of the week acid into 100 mL beaker
titrated this solution of 0.21 M NaOH
moles of weak acid = 3.75*10^-3
moles of NaOH = moles of week acid
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