Question

Sebanyak3.16 gr (ch3coo)2ca mr=158
Hit phnya jika dilarutkan dalam air 200ml
Ka=2x10^-5

Answer 1

H+=√ka.m = √2×10-4.1×10-4 =√2×10-9=1.4×10-4,5 pH=4.5-log1.4=4,5-0.14=4.36 Kalo gak slh sperti ini

Answer 2

Hidrolisis Garam

mol (CH₃COO)₂Ca = massa/Mr
mol (CH₃COO)₂Ca  = 3,16/158
mol (CH₃COO)₂Ca  = 0,02 mol

(CH₃COO)₂Ca  ⇆ 2CH₃COO⁻ + Ca²⁺
      0,02                   0,04              0,02


M CH₃COO⁻ = mol/Volume
M CH₃COO⁻ = 0,04/0,2
M CH₃COO⁻ = 0,2 M

[OH⁻] = √Kw/Ka • M anion garam
[OH⁻] = √10⁻¹⁴/2 x 10⁻⁵ • 0,2
[OH⁻] = √10⁻¹⁰
[OH⁻] = 10⁻⁵

pOH = -log [OH⁻]
pOH = -log [10⁻⁵]
pOH = 5

pH = 14 - pOH
pH = 14 - 5
pH = 9

Jadi, pH larutan adalah 9