2) 50 ml NH3 0,2m +100ml NH4BR 0,1M (ka=10-5)
Jawaban:
1.pH = 5 - log(3)
2.pH = 9
Penjelasan:
1. Diketahui:
Ka = 10^-5
[HCOOH] = 0,4 M
V HCOOH = 100mL
n = 10 mmol
[NaOH] = 0,1 M
V NaOH = 100mL
n = 10 mmol
HCOOH + NaOH ----> HCOONa + H2O
M: 40mmol. 10mmol. ----. ----
B:. -10mmol. -10mmol. +10mmol. + 10mmol
------------------------------------------------------------------+
S: 30mmol -----------. 10mmol. 10mmol.
Rumus larutan Penyangga asam lemah
[H+] = Ka x ([a]/[b]) = Ka x (a/b)
[H+] = 10^-5 x (30/10)
[H+] = 3 x 10^-5
pH = -log([H+])
pH = -log(3x10^-5)
pH = 5 - log(3)
2. Diketahui:
[NH3] = 0,2 M
v = 50mL
n = 10mmol
[NH4Br] = 0,1 M
v = 100mL
n = 10mmol
Rumus larutan penyangga basa lemah
[OH-] = Kb x ([b]/[a] = Kb x (b/a)
[OH-] = 10^-5 x (10/10)
[OH-] = 10^-5
pOH = 5
pH = 9
[answer.2.content]
