PH 100 mL larutan NH3 0,4 M adalah... (Kb NH3 = 9 x 10^-5)

A. 3-log 6
B. 5-log 3,6
C. 9 log 3,6
D. 11 log 6
E. 10 log 3,6

PH 100 mL larutan NH3 0,4 M adalah… (Kb NH3 = 9 x 10^-5)

Jawaban Terkonfirmasi

(OH⁻) = $sqrt{kb . cb}$
= $sqrt{9.10^{-5} . 0.4}$
= 6.10⁻³

pOH = – log (OH⁻)
= – log 6.10⁻³
= 3 – log 6

pH = 14 – (3 – log 6)
= 11 + log 6

Jawaban Terkonfirmasi

[OH-]= $sqrt{kb.Mb}$
= $sqrt{9.10^-5.4.10^-1}$
= $sqrt{36.10^-6}$
=6.10^-3
poh=-log [oh]
= -log[6.10^-3]
=3-log 6
ph=14-poh
=14-[3-log 6]
= 14-3+log 6
= 11+log 6