Berapa gramkah Ba(OH)2 yang dibutuhkan untuk membuat larutan pH =13+log 2 sebanyak 200mL larutan?

Materi : Asam Basa

pH = 13 + log 2 maka pOH = 1 – log 2

[OH-] = 0.2 M

valensi basa Ba(OH)2 = 2

[Ba(OH)2] = 0.1 M

Ba(OH)2 yg dilarutkan dlm 200 mL:
200/1000 × 0.1 × 171 = 3.42 gram

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Asam Basa
Kelas 11

pOH = pKw – pH
= 14 – (13+ log 2)
= 1 – log 2

pOH = -log [OH-]
1-log 2 = -log [OH-]
log 2 – 1 = log [OH-]
[OH-] = 1/5 = 0,2

[OH-] = M x valensi basa
0,2 = M x 2
M = 0,1 M

n = M x V
= 0,1 x 0,2 L
= 0,02 mol

Mr = 137 + 2(16+1)
= 171
m = n x Mr
= 0,02 . 171
= 3,42 gr