Q.20 how come it's 12.5 ?!
Okay from the First part of the Question we know that in the Titration
0.25 moldm-3 of NaOH and
0.125 moldm-3 of Ethanoic Acid !!
so What I did is I found the Ratio as follows => Acid : Base
0.125 : 0.25 divide by 0.125 for both
I get a Ratio of =>
1 : 2 And from the part (v) we get to know that
25 cm3 of the ACID is USED UP !!! .. so to find the Volume at the End point (Where the salt/product is given indicating that the ACID is completely KILLED) is
1 : 2 25 : X Cross multiply and You'll find that the Volume of the BASE is =>
12.5 cm3I simply used my common sense and it worked
if it doesn't work Use the formula from the CIE board (not in the syllabus) ;
(C1V1) acid = (C2V2)base.... and act smart