right you either go in to school and ask your chemistry or do it the messy way; get 11.30ml of the 0.10m NaOH(aq) with 8.00ml of the 0.10m HCI(aq) with 8.00ml of the 0.10m HC2H302(aq). mix them all together then look at the colour an there is you pH solution for that question.
by the way im only in year 8 at school.-you should know this
this is not done in lab. it's supposed to be calculated. people, like you, fail.
Doing it in the lab IS one way of solving some of these problems and lab work is quite important; however, one must ALSO know how to calculate the answers. A good chemist must know how to do both; that is, must be good in the lab as well as in theory. In this case, the problem is to be solved by calculation.
This is a buffer solution disguised so it's a little harder to recognize.
mols NaOH = L x M = 0.01130 x 0.100 M = ??
mols HCl = L x M = 0.00800 x 0.100 M = ??
Since the equation for the strong base and strong base is 1:1; i.e.,
NaOH + HCl ==> NaCl + H2O
Just subtracting mols NaOH - mols HCl will give you the amount of NaOH remaining (since it is the reagent in excess). THEN, you just have an equation of a strong base with a weak acid; i.e.,
NaOH + HC2H3O2 ==> NaC2H3O2 + H2O
which produces a buffer with the salt and the weak acid.
Calculate mols NaOH left from the first reaction. Calculate mols HC2H3O2 added. Determine the amount of HC2H3O2 left after the NaOH reaction and the amount of C2H3O2^- ion produced. You now have a weak acid and its conjugate base (HC2H3O2/C2H3O2^-) and you can use the Henderson-Hasselbalch equation.
pH = pKa + log(base/acid)
Post your work if you get stuck.
how the hell do u do this. just give me thanswER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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