0.005 mol of Ba OH 2 is dissolved in 100 mL of water. Assuming complete ionisation of Ba OH 2 the pOH of the solution will be

Question

0.005 mol of Ba OH 2 is dissolved in 100 mL of water. Assuming complete ionisation of Ba OH 2 the pOH of the solution will be, 1, 5, 2, 14

MARKS App · Accepted Answer

100 mL of solution contain 0.005 moles of Ba OH 2 ∴ 1 L 1000 mL of solution contain = 10 × 0 .005 mole of Ba OH 2 Concentration of Ba OH 2 (i.e. moles on litres) = 0 .05 M Each Ba OH 2 give 2OH − ions Thus, moles of OH − per L = 2 × 0.05 = 0.1 iii ∵ pOH = − log OH − pOH = − log 0 .1 = − log10 − 1 pOH = log 10 = 1

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