sebanyak 3,6 gram natrium benzoat dilarutkan kedalam air hingga volumenya 1 liter jika diketahui ka c6h5cooh = 6,5 x 10^-5 senyawa tersebut akan terhidrolisis

NaC6H5COO > Na^+ + C6H5COO^-

apakah reaksi tsb bersifat asam atau basa:

pada
C6H5COO^- + H2O > C6H5COOH + OH^- (basa)
mol NaC6H5COO = 3,6/144 = 0,025mol

[NaC6H5COO] = 0,025/1 = 0,025M

[OH^-] = akar Kw/Ka [C6H5COO^-]

[OH^-] = akar 10^-14/6,5x10^-5 (0,025)

[OH^-] = akar3,846 x 10^-12

[OH^-] = setara 2x10^-6 M

pOH = -log(2x10^-6)

pOH = 6 - log2
pOH = 5,7
pH = 14 - 5,7
PH = 8,3