Asked by Fai
The formula weight of an acid is 82. In a titration, 100cm3 of the solution of this acid containing 39g of this acid per litre were completely neutralized by 95 cm3 of aqueous naoh containing 40g of naoh per litre. What is the basicity of this acid?
Bob helps me
Bob helps me
Answers
Answered by
DrBob222
mols NaOH in solution = grams/formula weight = 40/40 = 1 mol.
M NaOH = 1 mol/L = 1M
mols acid = grams/formula weight = 39/82 = about 0.48 but you do it more accurately.
M base = mols/L = 0.48M
mols acid x #H = mols base
0.48M x 0.100L x #H = 1M x 0.095L
#H = (1M x 0.095)/(0.48 x 0.095)
#H = ?
M NaOH = 1 mol/L = 1M
mols acid = grams/formula weight = 39/82 = about 0.48 but you do it more accurately.
M base = mols/L = 0.48M
mols acid x #H = mols base
0.48M x 0.100L x #H = 1M x 0.095L
#H = (1M x 0.095)/(0.48 x 0.095)
#H = ?
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