HItunglah pH larutan berikut :
a. Larutan 500 mL amonia o,1 M ( Kb = 4 x 10^{-5} [/tex] )
b. Larutan 6 gram asetat dalam 200 mL air ( Ka =
)
c. Larutan 100 mL asam format ( α = 0,1% )
a. Larutan 500 mL amonia o,1 M ( Kb = 4 x 10^{-5} [/tex] )
b. Larutan 6 gram asetat dalam 200 mL air ( Ka =
c. Larutan 100 mL asam format ( α = 0,1% )
[OH-] = akar (kb x Mb) = akar (0,1 x 4 x 10^-5) = akar 4 x 10^-6 = 2 x 10^-3
pOH = - log 2 x10^-3 = 3 - log 2
pH = 14 - pOH = 11 + log 2
b. 6 gr as asetat dlm 200 mL air
cari konsentrasinya dahulu
M = gram/Mr x 1000/V = 6/60 x 1000/200 = 0,5 M
[H+] = akar (ka x Ma) = akar ( 0,5 x 10^-5) = akar 5 x 10^-6 = 10^-3 x akar 5
pH = - log H+ = 3 - log akar 5
c.100 ml asa formiat alpha 0,1
[H+] = alpha x Ma = ??????? x 0,1 =
pH = - log H+ = ??????
g ada konsentrasinya atau massa, gbs dicari, kurang mgkn pas ngetiknya :)
semoga bermanfaat :)
=akar 4x10⁻⁵ . 5x10⁻²
=akar 2x10⁻⁶
=akar 2 x 10⁻³
pOH = -log akar 2 x 10⁻³
= 3 - log akar 2
pH = 11+log akar 2