What is the [ OH -]concentration of a 0.04 M solution of CH 3 COONa ? [K a of CH 3 COONa =2× 10-5, log 2=0.30, log 5=0.69,100.7 ≅ 5]

Question

What is the [ OH -]concentration of a 0.04 M solution of CH 3 COONa ? [K a of CH 3 COONa =2× 10-5, log 2=0.30, log 5=0.69,100.7 ≅ 5] (A) 5× 10- 6 (B) 6× 10- 6 (C) 2× 10- 9 (D) 3× 10- 9

Tardigrade Admin · Accepted Answer

mathrmpK mathrma=- log mathrmKa=- log (2 × 10-5) =5- log 2=5-0.3=4.7 mathrmpH=(1/2)( mathrmpK mathrmw+ mathrmpK mathrma+ log mathrmC) p H=(1/2)[14+4.7+ log (0.04)] p H=8.63 pOH =14-8.63 ≅ 5.3 [ OH -]=5 × 10-6

Q. What is the $[O H^{-}]$ concentration of a $0.04 M$ solution of $C H_{3} COON a$ ? [ $K_{a}$ of $C H_{3} COON a = 2 \times 1 0^{- 5}, lo g 2 = 0.30, lo g 5 = 0.69, 1 0^{0.7} ≅ 5$ ]

$5 \times 1 0^{- 6}$

$6 \times 1 0^{- 6}$

$2 \times 1 0^{- 9}$

$3 \times 1 0^{- 9}$

$pK_{a} = - lo g K_{a} = - lo g (2 \times 1 0^{- 5})$
$= 5 - lo g 2 = 5 - 0.3 = 4.7$
$pH = \frac{1}{2} (pK_{w} + pK_{a} + lo g C)$
$p H = \frac{1}{2} [14 + 4.7 + lo g (0.04)]$
$p H = 8.63$
$pO H = 14 - 8.63 ≅ 5.3$
$[O H^{-}] = 5 \times 1 0^{- 6}$

Q. What is the [OH−]concentration of a 0.04M solution of CH3​COONa ? [Ka​ of CH3​COONa=2×10−5,log2=0.30,log5=0.69,100.7≅5]

5×10−6

6×10−6

2×10−9

3×10−9