Diketahui suatu larutan basa hasil pelarutan 0,4 gram NaOH (Ar Na = 23 , Ar O = 16 , Dan Ar H = 1 ) dalam 500 ml.Hitung lah A.Molaritas Dari Larutan Tersebut B.

a. M = gram/Mr x 1000/vol
= 0,4/40 x 1000/500
= 0,02 M

b. OH = M x Val = 0.02x1 = 0.02 = 2 x 10^-2
c. POH = 2 - log 2
d. pH = 14 - ( 2 - log 2) = 12 + log 2

semoga membantu ^^

Mapel; Kimia
Kelas: XI
Materi: Asam Basa

Jawab

Mr NaOH
Mr 23+16+1
Mr → 40gr/mol

V = gr/Mr ×1000/V
V= 0.4/40 × 1000/500
V= 0.01×2
V= 0.02M

Molaritas= 0,02M

NaOH→ Na+ + [OH-]
valensi basa 1

[OH-] = M×Vb
[OH-] = 0.02×1
[OH-] = 2×10-²

pOH = -log [OH-]
pOH = -log 2×10-2
pOH= 2-log2

pH= pKw-pOH
pH= 14-(2-log2)
pH= 12+log2
pH= 12+0.3
pH=12,3