pH

Submitted by Major on Sat, 03/22/2014 - 17:44
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Neki rastvor sadrži NaOH, C = 3,27×10-4 mol/dm3 i KOH, C = 1,23×10-2 mol/dm3. Kolika je pH vrednost rastvora?KW = 1,0008x10-14

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milicaradenkovic - Sat, 03/22/2014 - 20:15

c(OH-)=c1(OH-)+c2(OH-)

NaOH--------------------->Na+ + OH-

c1(OH-)=c(NaOH)=3.27*10(-4) mol/l

KOH--------------------->K+ +OH-

c2(OH-)=c(KOH)=1.23*10(-2) mol/l

c(OH-)=3.27*10(-4)+1.23*10(-2)=1.263*10(-2) mol/l

(H+)=Kw/c(OH-)=1.0008*10(-14)/1.263*10(-2)=7.92*10(-13) mol/l

pH= -log(H+)=12.1