Penjelasan:

mol HA = M . V = 0,1 . 200 = 20 mmol

mol NaOH = M . V = 0,1 . 100 = 10 mmol

a. pH campuran

HA + NaOH => NaA + H2O

mula-mula 20 10

bereaksi 10 10 10 10

sisa 10 - 10 10

karena asam lemah yang bersisa, maka terbentuk larutan penyangga asam.

[H^+] = Ka. (mol HA / mol NaA)

= 10^-4 . ( 10/10)

= 10^-4

pH = - log [H^+]

= - log (10^-4)

= 4

b. pH campuran setelah penambahan 10 mL HCl 0,1 M

mol HCl = M . V = 0,1 . 10 = 1 mmol

NaA + HCl => NaCl + HA

mula-mula 10 1 10

bereaksi 1 1 1 1

sisa 9 - 1 11

[H^+] = Ka. (mol HA / mol NaA)

= 10^-4 . ( 2/9)

= 1,2 . 10^-4

pH = - log [H^+]

= - log (1,2 . 10^-4)

= 4 - log 1,2

c. pH campurab setelah penambahan 10 mL NaOH 0,05 M

mol NaOH = M . V = 0,05 . 10 = 0,5 mmol

NaOH + HA => NaA + H2O

mula-mula 0,5 10 10

bereaksi 0,5 0,5 0,5 0,5

sisa - 9,5 10,5 0,5

[H^+] = Ka. (mol HA / mol NaA)

= 10^-4 . ( 9,5/10,5)

= 9 . 10^-5

pH = - log [H^+]

= - log (9 . 10^-5)

= 5 - log 9