200 ml larutan hcl 0.1 m dicampurkan dengan 200ml laruran Ba(OH)2 0,005m. Ph larutan tersebut.berapa? tolong yah kak ^^
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lindayuliyanti17
Pertanyaan
200 ml larutan hcl 0.1 m dicampurkan dengan 200ml laruran Ba(OH)2 0,005m. Ph larutan tersebut.berapa? tolong yah kak ^^
1 Jawaban
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1. Jawaban whongaliem
mol HCl = 200 ml x 0,1 M
= 20 m mol
mol Ba(OH)2 = 200 ml x 0,005 M
= 1 m mol
HCl ⇒ (H+) + Cl -
jumlah mol H+ = 1/1 x mol HCl
= 1 x 20 m mol
= 20 mmol
Ba(OH)2 ⇒ 2.Ba+ + OH-
jumlah mol OH- = 1/1 x mol Ba(OH)2
= 1 x 1
= 1 mmol
ion H+ berlebih = (20 - 1) m mol
= 19 mmol
[H+] = (sisa mol H+) / Vol campura
= 1 mmol / (200 + 200)ml
= 1/400 M
= 25 x 10^-4
PH = - log [H+]
= - log (25 x 10^-4)
= 4 - log 25
= 4 - 2.log 5