Question

Bantu please no 15,17,18

Answer 1

15. mmol CH3COOH = 400 x 0,5 = 200 mmol
mmol CH3COONa = 100 x 0,5 = 50 mmol
[H+] = Ka x mol asam/mol garam
[H+] = 10⁻⁵ x 200/50
[H+] = 4x10⁻⁵
pH = - log 4x10⁻⁵
pH = 5 - log 4

17. mmol CH3COOH = 400 x 0,2 = 80 mmol
mmol Ba(OH)2 = 200 x 0,1 = 20 mmol
2 CH3COOH + Ba(OH)2 => (CH3COO)2Ba + 2 H2O
a :  80              20
b :  40              20                    20                  40
s :  40               -                      20                  40      

[H+] = Ka x mol asam/mol garam
[H+] = 2 x10⁻⁵ x 40/20
[H+] = 4 x10⁻⁵
pH = - log 4 x10⁻⁵
pH = 5 - log 4