18L of 2M Ba(OH)2 are needed to titrate a 4L solution of 6M H3PO4 .
Given ,
volume of H3PO4 = 4liters
the reaction of H3PO4 , dissociate into H+ and PO4^3- .
H3PO4 → 3H+ + PO4^3-
Thus , the concentration of H+ , [H+] = 3 ×6 = 18M
the reaction Ba(OH)2 , dissociate into Ba^2+ + 2OH-
Ba(OH)2 →Ba^2+ + 2OH-
thus the concentration of OH- , [OH-] = 2×2 = 4M
According to equivalence ,
M1×V1 = M2×V2
18×4× = 4 ×V2
V2 = 18 liters
Hence , 18 liters of 2M Ba(OH)2 are needed to titrate 4L solution of 6M H3PO4 .
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