25cm3 of 0.10 mol dm-3 acrylic acid(C3H402-monobasic acid) was titrated against 0.10 mol dm-3 of barium hydroxide. For this titration. calculate the volume of Ba(OH)2 that must be added to produce a buffer solution of the best buffer capacity. ans:6.25cm3
i think need the eqn (2C3H402= Ba(OH)2) to solve the qn. but i dont understand how this eqn comes about.
Originally posted by lin2:25cm3 of 0.10 mol dm-3 acrylic acid(C3H402-monobasic acid) was titrated against 0.10 mol dm-3 of barium hydroxide. For this titration. calculate the volume of Ba(OH)2 that must be added to produce a bsuffer solution of the best buffer capacity. ans:6.25cm3
i think need the eqn (2C3H402= Ba(OH)2) to solve the qn. but i dont understand how this eqn comes about.
At maximum buffer capacity, molarity of weak acid = molarity of its conjugate base, which occurs at half-neutralization. Since barium hydroxide is a diprotic alkali, consider the molarity of hydroxide ions to be 0.2 mol/dm3.
Moles of monoprotic acrylic acid present = 2.5 x 10^-3
Hence, for half neutralization to occur to achieve molarity of weak acid = molarity of its conjugate base, we need 1.25 x 10^-3 moles of hydroxide ions.
Hence volume of barium hydroxide required = moles / molarity = (1.25 x 10^-3 ) / (0.2) = 6.25 x 10^-3 dm3 = 6.25 cm3
i dont understand why u can just take o.1 x2 to get conc of OH- ions.. just because its a diprotic alkali?
Originally posted by lin2:i dont understand why u can just take o.1 x2 to get conc of OH- ions.. just because its a diprotic alkali?
Yes, that's right. 1 mol/dm3 of Ba(OH)2 = 2 mol/dm3 of OH-.
erm, since molarity of acid need to be equal to the molarity of conjugate base, and the number of moles of acid=2.5x10-3, the number of moles of conjugate base should be (2 X 2.5x10-3?(since its diprotic) H+=20H-
does the 'molarity' refers to concentration?
Originally posted by lin2:erm, since molarity of acid need to be equal to the molarity of conjugate base, and the number of moles of acid=2.5x10-3, the number of moles of conjugate base should be (2 X 2.5x10-3?(since its diprotic) H+=20H-
does the 'molarity' refers to concentration?
Molarity = molar concentration. It's better to use the more unambiguous term "molarity" since concentration can be mol/dm3, g/dm3, etc.
At initial, we have 2.5 x 10^-3 moles of the weak acid.
We must add 1.25 x 10^-3 moles of hydroxide ions, so that upon acid-base neutralization, we end up with only 1.25 x 10^-3 moles of the weak acid and 1.25 x 10^-3 moles of its conjugate base. This is point of maximum buffer capacity.
An ICF table (Initial, Change, Final) would show this clearly, and its recommended that students use ICF tables together with ICE tables (Initial, Change, Equilibrium).
ICF = human action; ICE = universe action.
ICF = single arow; ICE = double half-arrow.
ICF = moles, moles, moles; ICE = molarity, molarity, molarity.
ICF = usually no algebra used; ICE = algebra usually used.
Some JCs teach only ICF, some teach only ICE, some do not teach either, some teach both. Cambridge doesn't care whether you use either, neither or both, as long as you get the correct answer. These tables are simply optional tools to help you.
If you're interested to use them (eg. if you're struggling with this topic), you can search them out on the internet, or look up undergrad textbooks like Raymond Chang's "Chemistry", or come for my tuition, etc.
ok.. so i need to divide the 2.5 x 10-3 moles by 2 to get the same amount of weak acid and conjugate base?
i think i get it alr. thanks!