Question

In '$x$' ml $0.3N$ $HCl$, addition of $200ml$ distilled water or addition of $100ml$ $0.1N$ $NaOH$, gives same final acid strength. Determine '$x$'.

Answer 1

Case $I$:
When $200ml$ distilled water is added to $xmL$ solution,
$x\times 0.3=(x+200)\times y$
Here, $y=$ final normality or concentration
$\therefore$ $y=\frac{0.3x}{(x+200)}$
Case $II$:
Number of equivalents of $HCl=\frac{0.3\times x}{1000}$
Number of equivalents of $NaOH=\frac{100\times 0.1}{1000}=0.01$
Number of equivalents of $HCl$ remained after addition of $NaOH=\frac{0.3x}{1000}-0.01$
$\therefore$ Concentration $=\frac{\{\frac{0.3x}{1000}-0.01\}}{100+x}\times 1000$
As final acid strength is same in both cases
$\frac{\{\frac{0.3x}{1000}-0.01\}1000}{100+x}=\frac{0.3x}{200+x}$
or, $\frac{0.3x-10}{100+x}=\frac{0.3x}{(200+x)}$
or, $(0.3x-10)\times (200+x)=\left(0.3x\right)(100+x)$
or, $60x-2000+0.3x^2-10x=30x+0.3x^2$
or, $20x=2000$
or, $x=100ml$