1.

Volume of 0.5 M Ba (OH)_(2) require to neutralize 100 ml 0.8 M H_(3) PO_(3) is :

Answer»

`100 ML`
`240 ml`
`160 ml`
`120 ml`

Solution :`Ba(OH)_(2) rarr n = (0.5 xx V mL)`
`H_(3)PO_(3) rarr n = (0.8 xx 100 mL)`
`H_(3)PO_(3)` is a diabasic acid
Equal moles of `Ba(OH)_(2)` and `H_(3)PO_(3)` are needed
`n_(Ba (OH)_(2)) = n_(H_(3)PO_(3)`
`M_(1)V_(1) = M_(2) V_(2)`
`0.5 xx (V)/(1000) = 0.8 xx (100)/(1000)`
`V = (8 xx 100)/(5) = 6 mL`


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