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Solution :

In this case, strong acids and strong bases are mixed. Equal amount of mili equivalents of acid and base will neutralise. Concentrations of `HCl` and `Ba(OH)_(2)` are in molarties, they have to the converted to normalitites as concentration of `H_(2) SO_(4)` is given in normality. <br> `:.` Total acids `= 50 xx 0.2 xx 1 + 50 xx 0.2 xx 1` ltbgt `= 10 + 10 = 20 mEq` <br> Total base `= 200 xx 0.2 xx 2` (n factor) <br> `= 80 mEq` <br> Since 20 mEq of acid would neutralise 20 mEq of base, thus, <br> Base left `= 80 - 20 = 60 mEq` <br> Total volume of solution `= 1 L = 1000 mL` <br> So, final concentration of solution `= (mEq)/(mL) = (60)/(1000) = 0.06 N` <br> Final resulting mixture of solution is basic, since 60 mEq of strong base is left.