Massa NaOH (Mr= 40) yang harus dilarutkan untuk membuat larutan NaOH yang mempunya pH=12 sebanyak 500 mL yaitu ... g ?

PH = 12
V = 500 ml
Mr = 40

pOH = pKW - pH
pOH = 14 - 12
pOH = 2

pOH = 2
-log OH- = 2. log 10
Log OH- = - 2. Log 10
Log OH' = log 10^-2
OH- = 10^-2

OH- = 1.Mb
10^-2 = Mb
Mb = 10^-2

Mb = (m/Mr)( 1000/V)
10^-2 = (m/40)(1000/500)
10^-2 = (m/40) 2
10^-2 = m/20
m =( 10^-2) 20
m = (10^-2)(2.10)
m = 2.10^-2 g atau 0,2 g

[tex]ph = 12 \\ ( h +) = {10}^{ - 12} \\ (oh - ) = {10}^{ - 2} \\ (oh - ) = m \times valensi \\ {10}^{ - 2} = m \times 1 \\ {10}^{ - 2} = \frac{x}{40} \times \frac{1000}{500} \\ x = 0.2 \: gram[/tex]
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