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50 mL of 1 M HCl, 100 mL of 0.5 M HNO(3)...

`50 mL` of `1 M HCl, 100 mL` of `0.5 M HNO_(3)`, and `x mL` of `5 M H_(2)SO_(4)` are mixed together and the total volume is made upto `1.0 L` with water. `100 mL` of this solution exactly neutralises `10 mL` of `M//3 Al_(2) (CO_(3))_(3)`. Calculate the value of `x`.

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The correct Answer is:
A
Total mEq of acid `= 50 xx 1 + 100 xx 0.5 + x xx 5 xx 2` (`n` factor )
`= (100 + 10x) mEq`
`= ((100 + 10 x))/(100 mL) N`
`N_(1) V_(1)` (Acid) `= N_(1) V_(1) [Al_(2) (CO_(3)^(2-))_(3)]` (Total charge = 6)
`(n = 6)`
`:. ((100 + 10 x))/(100 mL) N xx 100 mL = 10 mL xx (1)/(3) xx 6`
`(100 + 10x) = 200`
`:. x = 10 mL`
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