The pOH of 0.0005 M sulphuric acid is

Question

The pOH of 0.0005 M sulphuric acid is (A) 5 (B) 3 (C) 11 (D) 12

Tardigrade Admin · Accepted Answer

H 2 SO 4 leftharpoons 2 H ++ SO 42- So, [ H +]=2 × 0 ⋅ 0005=1 × 10-3 M pH =- log [ H +]=- log (1 × 10-3) pH =3 ∵ pH + pOH =14 3+ pOH =14 pOH =14-3 pOH =14 pOH =11

Q. The pOH of $0.0005 M$ sulphuric acid is

5

3

11

12

$H_{2} S O_{4} ⇌ 2 H^{+} + S O_{4}^{2 -}$
So, $[H^{+}] = 2 \times 0 \cdot 0005 = 1 \times 1 0^{- 3} M$
$p H = - lo g [H^{+}] = - lo g (1 \times 1 0^{- 3})$
$p H = 3$
$∵ p H + pO H = 14$
$3 + pO H = 14$
$pO H = 14 - 3$
$pO H = 14$
$pO H = 11$

Q. The pOH of 0.0005M sulphuric acid is

5

3

11

12

H2​SO4​⇌2H++SO42−​ So, [H+]=2×0⋅0005=1×10−3M pH=−log[H+]=−log(1×10−3) pH=3 ∵pH+pOH=14 3+pOH=14 pOH=14−3 pOH=14 pOH=11

Q. The pOH of $0.0005 M$ sulphuric acid is

$H_{2} S O_{4} ⇌ 2 H^{+} + S O_{4}^{2 -}$
So, $[H^{+}] = 2 \times 0 \cdot 0005 = 1 \times 1 0^{- 3} M$
$p H = - lo g [H^{+}] = - lo g (1 \times 1 0^{- 3})$
$p H = 3$
$∵ p H + pO H = 14$
$3 + pO H = 14$
$pO H = 14 - 3$
$pO H = 14$
$pO H = 11$