Question

Tolong Jawab Lah Gays Bagi yang Bisa tolong

Answer 1

NaOH → Na^[+] + OH^[-]

0,02 0,02 0,02

pH = 14 - log pOH
pH = 14 - log 0,02
pH = 12 + log 2

[H^+] = gr/Mr × 1/L
[H^+] = 3,7/74 × 1/10
[H^+] = 0,005 M

pH = - log [H^+]
pH = - log 0,005
pH = 3 - log 5