In 'x' mL 0.3 N HCl, addition of 200 mL distilled water or addition of 100 mL 0.1 N NaOH, gives same final acid strength. Determine 'x'.
Case I
When 200 mL distilled water is added to x mL solution.
x × 0.3 = ( x + 200) × y
Here, y = final normality or concentration
y =
Case II
Number of equivalents of HCl =
Number of equivalents of NaOH =
Number of equivalents of HCl remained after addition of NaOH =
As final acid strength is same in both cases
or,
or, (0.3x - 10) × (200 + x) = (0.3x)(100 + x)
or, 60x - 2000 + 0.3x2 - 10x = 30x + 0.3x2
or, 20x = 2000
or, x = 100 mL
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