If a 19.5 mL sample of 1.5 M solution of each of the following acids is...
If a 19.5 mL sample of 1.5 M solution of each of the following acids is reacted with 0.80 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?
19.5 mL of H2SO4 titrated with 0.80
M NaOH:
volume of NaOH ______ mL
total volume______ mL
19.5 mL of HC2H3O2 titrated with 0.80 M NaOH:
volume of NaOH_____ mL
total volume ____mL
19.5 mL of HCl titrated with 0.80 M NaOH:
volume of NaOH_______ mL
total volume ______mL
Homework Answers
H2SO4(aq) + 2NaOH(aq) --------------> Na2SO4(aq) + 2H2O(l)
1 mole 2moles
H2So4 NaOH
M1 = 1.5M M2 = 0.80M
V1 = 19.5ml V2 =
n1 = 1 n2 = 2
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 1.5*19.5*2/(0.80*1)
= 73.125ml
volume of NaOH = 73.125ml
total volume = 73.125 + 19.5 = 92.625ml
CH3COOH(aq) + NaOH(aq) --------------> CH3COONa(aq) + H2O(l)
1 mole 1moles
CH3COOH NaOH
M1 = 1.5M M2 = 0.80M
V1 = 19.5ml V2 =
n1 = 1 n2 = 1
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 1.5*19.5*1/(0.80*1)
= 36.5625ml
volume of NaOH = 36.5625ml
total volume = 36.5625 + 19.5 = 56.0625ml
HCl(aq) + NaOH(aq) -------------->NaCl(aq) + H2O(l)
1 mole 1moles
HCl NaOH
M1 = 1.5M M2 = 0.80M
V1 = 19.5ml V2 =
n1 = 1 n2 = 1
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 1.5*19.5*1/(0.80*1)
= 36.5625ml
volume of NaOH = 36.5625ml
total volume = 36.5625 + 19.5 = 56.0625ml
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