Sebanyak 25ml KOH 0,2 M dicampur dengan CH3COOH 0,2 M (Ka = 10^-5) dan ph larutan menjadi 9, maka volume asam asetat yang dicampurkan adalah?
Amaldoft
Diket dan dit: *n KOH = 25 mL x 0,2 M = 5 mmol *n CH3COOH = 0,2 M x v mL = 0,2V *Ka CH3COOH = 10^-5 *pH = 9 --> [H+] = 10^-9 *Volume CH3COOH = ... mL ?
Pertama, buat reaksi supaya mendapatkan garam: KOH + CH3COOH --> CH3COOK + H2O m 10 0,2V - - b -10 -10 +10 +10 s - 0,2V-10 +10
Bersisa asam lemah (CH3COOH) dan garamnya (CH3COOK), maka sistem buffer terjadi di sini, sehingga: [H+] = Ka x CH3COOH/CH3COOK 10^-9 = 10^-5 x (0,2V-10) / 10 10^-4 x 10 = 0,2V-10 0,004 = 0,2V - 10 0,2V = 10.004 0,2V = 10 V = 50 mL (B)
*n KOH = 25 mL x 0,2 M = 5 mmol
*n CH3COOH = 0,2 M x v mL = 0,2V
*Ka CH3COOH = 10^-5
*pH = 9 --> [H+] = 10^-9
*Volume CH3COOH = ... mL ?
Pertama, buat reaksi supaya mendapatkan garam:
KOH + CH3COOH --> CH3COOK + H2O
m 10 0,2V - -
b -10 -10 +10 +10
s - 0,2V-10 +10
Bersisa asam lemah (CH3COOH) dan garamnya (CH3COOK), maka sistem buffer terjadi di sini, sehingga:
[H+] = Ka x CH3COOH/CH3COOK
10^-9 = 10^-5 x (0,2V-10) / 10
10^-4 x 10 = 0,2V-10
0,004 = 0,2V - 10
0,2V = 10.004
0,2V = 10
V = 50 mL (B)