Question

Un volum de 2L solutie de soda caustica contine 8 g hidroxid de sodiu ,calculati pH-ul solutiei

Answer 1

V=2l
m=8g
pH=?
MNaOH=23+16+1=40 u.a.m
Cm=md/M*V=8/2*40=0,1 l
pOH=10(-1)
pH=10(-13)

Answer 2

Vs = 2L
m = 8g NaOH
pH = ?

Cm = md/M*Vs(L) = 8/(40*2) = 0,1 moli/L

Sau

n = m/M = 8/40 = 0,2 moli NaOH

C = n/Vs = 0,2/2 = 0,1 moli/L NaOH

NaOH -> B.T (bază tare) ==> [NaOH]=C=[OH-] = 10^(-1) moli/L

Kw = [OH-]*[H3O+] = 10^(-14) (moli/L)^2 ==> [H3O+] = 10^(-14) / 10^(-1) = 10^(-13) moli/L

pH = -lg[H3O+] = -lg10^(-13) = 13

Sau

pOH = -lg[OH-] = -lg10^(-1) = 1

pH + pOH = 14 => pH = 14-1 = 13