100 mL KOH 0,02 M
100 mL H2SO4 0,02 M
a.) pH masing-masing larutan sebelum direaksikan
*pH KOH
[OH^-] = b × Mb
[OH^-] = 1 × 0,02
[OH^-] = 0,02
[OH^-] = 2 × 10^-2
pOH = - log [OH^-]
pOH = - log 2 × 10^-2
pOH = 2 - log 2
pH = 14 - pOH
pH = 14 - (2 - log 2)
pH = 12 + log 2
*pH H2SO4
[H^+] = a × Ma
[H^+] = 2 × 0,02
[H^+] = 0,04
[H^+] = 4 × 10^-2
pH = - log [H^+]
pH = - log 4 × 10^-2
pH = 2 - log 4
b.) pH campuran
*n KOH = M × V
n KOH = 0,02 × 100
n KOH = 2 mmol
*n H2SO4 = M × V
n H2SO4 = 0,02 × 100
n H2SO4 = 2 mmol
#persamaan reaksi
... 2 KOH + H2SO4 => K2SO4 + 2 H2O
M : ... 2 .......... 2
R : .... 2 .......... 1
_____________---
S : ... --- .......... 1 ............... 1 ............ 2
yang sisa H2SO4
M = n/Vtotal
M = 1/(100+100)
M = 1/200
M = 0,005
[H^+] = a × Ma
[H^+] = 2 × 0,005
[H^+] = 0,01
[H^+] = 10^-2
pH = - log [H^+]
pH = - log 10^-2
pH = 2
c.) Sifat larutan hasil pencampuran adalah asam.
Materi : Titrasi Asam Basa [answer.2.content]
