Jawaban:

40 mL NH4OH 0,15 M + 20 mL HCl 0,3 M

Kb = 10^-5

pH ... ?

n NH4OH = M × V

n NH4OH = 0,15 × 40

n NH4OH = 6 mmol

n HCl = M × V

n HCl = 0,3 × 20

n HCl = 6 mmol

#Reaksi

... NH4OH + HCl => NH4Cl + H2O

M : ... 6 ......... 6

R : ... 6 ......... 6

_____________---

S : ... --- ....... --- ........... 6 ......... 6

M = n/V total

M = 6 / (40+20)

M = 6 / 60

M = 0,1

Garam NH4Cl bersifat asam

[H^+] = √( Kw/Kb × M )

[H^+] = √( 10^-14/10^-5 × 0,1 )

[H^+] = √( 10^-9 × 0,1 )

[H^+] = √( 10^-10 )

[H^+] = 10^-5

pH = - log [H^+]

pH = - log 10^-5

pH = 5

Penjelasan:

maaf kalo salah