100 ml CH3COOH 0,1 M + 100 ml CH3COONa 0,1 M Hitunglah jika dilarutkan hingga volume 500 ml
Amaldoft
Diket dan dit: *n CH3COOH = n CH3COONa = 100 mL x 0,1 M = 10 mmol *Ka CH3COOH = 10^-5 *Volume pair = 500 mL *pH ?
Jika ada kata "air", artinya dilakukan pengenceran. Pengenceran dalam buffer tidak merubah nilai pH sedikit pun. Mari buktikan dari pH awal hingga pengenceran dulu ya:
A. pH Buffer awal (sebelum pengenceran) [H+] = Ka x n CH3COOH/nCH3COONa = 10^-5 x 10/10 = 10^-5 pH = 5
B. pH pengenceran 500 mL air --> [CH3COONa] = [CH3COOH] = 10 mmol/100 mL + 100 mL + 500 mL = 0,014 molar --> [H+] = Ka x [CH3COOH]/[CH3COONa] = 10^-5 x 0,014/0,014 = 10^-5 pH = 5
Terbukti, bukan? :)
2 votes Thanks 1
alequa
itu 10 mmol/100 mL + 100 mL + 500 mL kok jdi 0,014 nya bknnya 600,1 ?
#maaf merepotkan
*n CH3COOH = n CH3COONa = 100 mL x 0,1 M = 10 mmol
*Ka CH3COOH = 10^-5
*Volume pair = 500 mL
*pH ?
Jika ada kata "air", artinya dilakukan pengenceran. Pengenceran dalam buffer tidak merubah nilai pH sedikit pun. Mari buktikan dari pH awal hingga pengenceran dulu ya:
A. pH Buffer awal (sebelum pengenceran)
[H+] = Ka x n CH3COOH/nCH3COONa
= 10^-5 x 10/10
= 10^-5
pH = 5
B. pH pengenceran 500 mL air
--> [CH3COONa] = [CH3COOH] = 10 mmol/100 mL + 100 mL + 500 mL
= 0,014 molar
--> [H+] = Ka x [CH3COOH]/[CH3COONa]
= 10^-5 x 0,014/0,014
= 10^-5
pH = 5
Terbukti, bukan? :)