Sebanyak 3.42 gram ba(oh) (mr=171)dilarutkan dlm 250 ml larutan,harga ph larutan tersebut yaitu

Sebanyak 3.42 gram ba(oh) (mr=171)dilarutkan dlm 250 ml larutan,harga ph larutan tersebut yaitu

Gr = 3,42g
Mr = 171
v = 250

Kita cari konsentrasinya dulu ya

$M = frac{gr}{Mr} .frac{1000}{V} \ M = frac{3,42}{171} . frac{1000}{250} \ M = frac{3,42}{171} . 4 \ M= frac{13,68}{171} \ M = 0,08 M$

Sekarang cari konsentrasi [OH-] dan pOH

$[0H^{-}]=M.i = 0,08 . 2 = 16 . 10 ^{-2} \ pOH = - log [0H^{-}] = - log (16 . 10 ^{-2})=2-log16$

Sekarang cari pH

$pH = 14 - pOH = 14 - (2 - log 16) = 12 + log 16$

Jadi pH = 12 + log 16