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−<br />

[ CH COOH][ OH ] +<br />

Kh = 3<br />

.<br />

[ H]<br />

− +<br />

[ CH3COO ] [ H]<br />

[ CH ]<br />

=<br />

3COOH<br />

[ CH3COO ][ H ]<br />

Ingat reaksi ionisasi CH3COOH: CH3COOH(aq) H + (aq) + CH3COO – (aq)<br />

+ −<br />

[ H ] [ CH3COO ] [ CH3COOH] Ka = ;<br />

[ CH3COOH] + − [ H ][ CH3COO ]<br />

Ingat juga:<br />

− + . [OH – ] [H + ] ............................................ (7)<br />

= 1<br />

K a<br />

K w = [OH – ] [H + ] ; sehingga per samaan (7) dapat ditulis:<br />

Kh = K<br />

K w .......................................................................................... (8)<br />

a<br />

Dari persamaan (6) dan persamaan (8), kalian dapat memperoleh persamaan<br />

berikut:<br />

[OH – ] = K<br />

K w<br />

b<br />

. [G]<br />

– log [OH – ] = – 1<br />

2 log K w<br />

Karena pK w<br />

pOH = – 1<br />

2 (pKw – pKa pH = pKw – pOH<br />

1<br />

– 2 log pK 1<br />

a + 2 log [G]<br />

– log [G])<br />

= 1<br />

2 (pKw + pKa + log [G])<br />

= 14, dan [G] = konsentrasi garam yang terhidrolisis, maka:<br />

pH = 1<br />

2 (14 + pKa + log [G])<br />

Keterangan:<br />

Kw : tetapan ionisasi air (10 –14 )<br />

Ka : tetapan ionisasi asam lemah<br />

[G] : konsentrasi ion garam yang terhidrolisis<br />

Contoh<br />

Berapa pH larutan yang terbentuk pada hidrolisis garam NaCN 0,01 M,<br />

jika diketahui Ka HCN = 1 10 –10 ?<br />

Penyelesaian:<br />

Diketahui:<br />

[NaCN] = 0,01 M<br />

Ka HCN = 1 10 –10<br />

Ditanyakan: pH larutan NaCN.<br />

Jawab:<br />

NaCN adalah garam dari asam lemah dan basa kuat. Jangan lupa yang<br />

kuat menentukan sifat garam dalam air. Jadi, garam bersifat basa, artinya<br />

memiliki ion OH – .<br />

Hidrolisis Garam<br />

209

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