Page 27 - E Modul Hidrolisis Garam baru

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Jawab :

Mol Ba(OH) 2 = Volume x Molaritas

= 0,01 L x 0,005 M
-5
= 5 x 10 mol

Mol HCN = Volume x Molaritas

= 0,01 L x 0,01 M

-4
= 1 x 10 mol

Ba(OH) 2(aq) + 2HCN (aq) Ba(CN) 2(aq) + 2H 2O (l)

-5
-5
Mula-mula : 5 x 10 mol 10 x 10 mol - -
-5 -5 -5 -5
Reaksi : 5 x 10 mol 10 x 10 mol 5 x 10 mol 10 x 10 mol
-5 -5
Sisa : - - 5 x 10 mol 10 x 10 mol

Mol Ba(CN) 2 = Ba(CN)2 mol Ba(OH) 2
Ba(OH)2

-5
1
-5
= x 5 x 10 mol = 5 x 10 mol
1

Konsentrasi Ba(CN) 2 = [Ba(CN) 2]

-3
= mol = 5 x 10 −5 mol = 2,5 x 10 M
volume total 5 x 10 −2 L