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

Options

A. `100 ml`
B. `240 ml`
C. `160 ml`
D. `120 ml`

Answer

C

Explanation

`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`