Jawaban:

0,3 M

Penjelasan:

asam asetat → asam lemah

NaOH → basa kuat

karena pH = 5 → larutan bersifat asam

Jadi larutan adalah buffer asam

mol asam = 100 ml x M asam asetat

= 100 M

mol basa = mol NaOH

= 100 ml x 0,2 M = 20 mmol

Persamaan reaksi:

CH3COOH + NaOH → CH3COONa + H2O

100 M mmol+ 20mmol

-20 mmol. -. 20mmol

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(100M -20) habis...........20 mmol

pH = 5

[H⁺] = 10⁻⁵

[H⁺] = Ka. mol asam/mol garam

10⁻⁵ = 2. 10⁻⁵.( 100M-20)/ 20

10⁻⁵. 20 = 2 .10⁻⁵.( 100M-20)

2. 10⁻⁴ = 2.10⁻³ M - 4. 10⁻⁴

6. 10⁻⁴ = 2.10⁻³ M

M = 0,3

Jadi molaritas asam asetat 0,3 M