Question

Calculate for 0.01 N solution of sodium acetate (i) Hydrolysis constant (ii) Degree of hydrolysis (iii) pH Given Ka of CH3COOH = 1.82 × 10–5

Answer 1

here is your answer

(OH-) from NaOH,a strong base =ch=0.1×10^-4=2.29×10^-6M pOH=5.64

pH=14-5.64=8.36

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Answer 2

Answer:

Explanation:

For     CH3COONa  +   H20 ⇄ CH3COOH  +  NaOH

initial           C                                     0                     0

after             C(1-h)                            Ch                      Ch

$K_{h}  = K_{w} /K_{a}  =\frac{ 10^{-14}}{1.9 10^-5 }   = 5.26 *10^{-10}$

h = $\sqrt{\frac{K_{h} }{C} } =\sqrt{\frac{5.26*10^{-10} }{0.01} }$

= $2.29*10^{-10}$

$OH^{-}$ from NaOH, a strong base = Ch = 0.01 × 2.29 × 10^–4

= 2.29 × 10–6 M pOH

= 5.64  pH

= 14 – 5.64

= 8.36