Ionic EquilibriumHard
Question
A 0.1 M sodium acetate solution was prepared. The Kh = 5.6 × 10-10
Options
A.The degree of hydrolysis is 7.48 × 10-5
B.The [OH-] concentration is 7.48 × 10-3 M
C.The [OH-] concentration is 7.48 × 10-6 M
D.The pH is approximately 8.88.
Solution
CH3COO- + H2O CH3COOH + OH-.
0.1 (1 - h) 0.1 h 0.1 h.
Kh =
= 0.1 h2 ⇒ 5.6 × 10-10 = 0.1 h2
⇒ h = 7.48 × 10-5 [∴ h <<<< 1]
[OH-] = ch = 7.48 × 10-5 × 10-1 = 7.48 × 10-6.
[H+] =
= 1.33 × 10-9.
⇒ pH = 8.8 approx.
0.1 (1 - h) 0.1 h 0.1 h.
Kh =
= 0.1 h2 ⇒ 5.6 × 10-10 = 0.1 h2 ⇒ h = 7.48 × 10-5 [∴ h <<<< 1]
[OH-] = ch = 7.48 × 10-5 × 10-1 = 7.48 × 10-6.
[H+] =
= 1.33 × 10-9.⇒ pH = 8.8 approx.
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