Determine the volume of the acid sample tested.?
Answer:
The normality of the acid solution = 1.15x10^-2 x3 (As the acid is H3X) ;
or = 3.45x10^-2N
For the titration ;
N1V1 = N2V2
Here N1 =0.1N, V1 = 16.8mL;
N2 = 3.45x10^-2N, V2 =?
or 0.1x16.8 = 3.45x10^-2 x V2 ;
or V2 = 0.1x16.8 /3.45x10^-2;
V2 = 48.696 mL or say 48.7 mL
Hence, the volume of acid tested = 48.7mL.
16.8 mL of LiOH solution is 0.0168 L.
The solution is 0.100 mol of LiOH per L.
So you have 0.00168 mol of LiOH that you've used.
3 LiOH + H3X = Li3X + 3 H2O (triple equivalence)
You have consumed one-third as much H3X on a mole basis.
0.00168 mol LiOH x (1 mol H3X / 3 mol LiOH) = 0.000560 mol of H3X.
0.000560 mol H3X x (1 L / 0.0115 mol) = 0.0487 L = 48.7 mL.
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