CH3COOH is 1 % ionised in its aqueous solution of 0.1 M strength. Its pOH will be

Question

CH3COOH is 1 % ionised in its aqueous solution of 0.1 M strength. Its pOH will be (A)  11  (B)  3  (C)  13  (D)  2

Tardigrade Admin · Accepted Answer

beginmatrix&CH3COOH+H2O&<=>&CH3COO-&+&H3O+ textInitial conc.&C&&0&&0 textconc. at equilibrium &C â Cα&&C α&&C α endmatrix Thus, [H+]=Cα=0.1×0.01=0.001 [α=1 %] ⇒ 10-3M ⇒ [OH-] =(10-14/10-3)=10-11 ⇒ pOH=11

Q. $C H_{3} COO H$ is $1%$ ionised in its aqueous solution of $0.1 M$ strength. Its $pO H$ will be

$11$

$3$

$13$

$2$

$Initial conc. conc. at equilibrium C H_{3} COO H + H_{2} O C C — C α C H_{3} CO O^{-} 0 C α + H_{3} O^{+} 0 C α$
Thus, $[H^{+}] = C α = 0.1 \times 0.01 = 0.001 [α = 1%]$
$\Rightarrow 1 0^{- 3} M$
$\Rightarrow [O H^{-}]$
$= \frac{1 0 ^{- 14}}{1 0 ^{- 3}} = 1 0^{- 11}$
$\Rightarrow pO H = 11$

Q. CH3​COOH is 1% ionised in its aqueous solution of 0.1M strength. Its pOH will be

11

3

13

2

Initial conc.conc. at equilibrium ​CH3​COOH+H2​OCC—Cα​​​CH3​COO−0Cα​+​H3​O+0Cα​ Thus, [H+]=Cα=0.1×0.01=0.001[α=1%] ⇒10−3M ⇒[OH−] =10−310−14​=10−11 ⇒pOH=11

Q. $C H_{3} COO H$ is $1%$ ionised in its aqueous solution of $0.1 M$ strength. Its $pO H$ will be

$11$

$3$

$13$

$2$

$Initial conc. conc. at equilibrium C H_{3} COO H + H_{2} O C C — C α C H_{3} CO O^{-} 0 C α + H_{3} O^{+} 0 C α$
Thus, $[H^{+}] = C α = 0.1 \times 0.01 = 0.001 [α = 1%]$
$\Rightarrow 1 0^{- 3} M$
$\Rightarrow [O H^{-}]$
$= \frac{1 0 ^{- 14}}{1 0 ^{- 3}} = 1 0^{- 11}$
$\Rightarrow pO H = 11$