A solution was prepared by dissolving 0.0005 mol of Ba(OH)2 in 100 mL of the solution. If the base is assumed to ionise completely, the pOH of the solution will be

Question

A solution was prepared by dissolving 0.0005 mol of Ba(OH)2 in 100 mL of the solution. If the base is assumed to ionise completely, the pOH of the solution will be (A) 12 (B) 10 (C) unpredictable (D) 2

Tardigrade Admin · Accepted Answer

Ba(OH)2 → Ba+2+2OH- Conc. of Ba(OH)2=0.0005×10=0.005 M OH- conc =2 × 0.005=0.01=10-2 pOH=-log[OH-]=-log 10-2=2

Q. A solution was prepared by dissolving 0.0005 mol of $B a (O H)_{2}$ in 100 mL of the solution. If the base is assumed to ionise completely, the pOH of the solution will be

12

10

unpredictable

2

$B a (O H)_{2} \to B a^{+ 2} + 2 O H^{-}$
Conc. of $B a (O H)_{2} = 0.0005 \times 10 = 0.005 M$
$O H^{-}$ conc $= 2 \times 0.005 = 0.01 = 1 0^{- 2}$
$pO H = - l o g [O H^{-}] = - l o g 1 0^{- 2} = 2$

Q. A solution was prepared by dissolving 0.0005 mol of Ba(OH)2​ in 100 mL of the solution. If the base is assumed to ionise completely, the pOH of the solution will be

12

10

unpredictable

2

Ba(OH)2​→Ba+2+2OH− Conc. of Ba(OH)2​=0.0005×10=0.005M OH− conc =2×0.005=0.01=10−2 pOH=−log[OH−]=−log10−2=2

Q. A solution was prepared by dissolving 0.0005 mol of $B a (O H)_{2}$ in 100 mL of the solution. If the base is assumed to ionise completely, the pOH of the solution will be

$B a (O H)_{2} \to B a^{+ 2} + 2 O H^{-}$
Conc. of $B a (O H)_{2} = 0.0005 \times 10 = 0.005 M$
$O H^{-}$ conc $= 2 \times 0.005 = 0.01 = 1 0^{- 2}$
$pO H = - l o g [O H^{-}] = - l o g 1 0^{- 2} = 2$