Oblicz wartość pH roztworu wodnego fenyloetyloaminy, jeżeli jego stężenie molowe wynosi 0,001 mol/dm.... Question from @GosiaBolis

December 2018 1 500 Report

Oblicz wartość pH roztworu wodnego fenyloetyloaminy, jeżeli jego stężenie molowe wynosi 0,001 mol/dm^3, a stała dysocjacji Kd=10^-9

StrawberryGum5 Należy tutaj zauważyć, że mamy do czynienia ze słabą zasadą

c = 0,001 mol/dm3
K = 10^-9
sprawdzam, czy mogę użyć gotowego wzoru
C/K = dużo większy od 400 więc mogę z niego skorzystać

K = alfa^2 x C
alfa = vK/C (pierwiastek z K /C)
alfa = 0,0316

[OH-] = alfa x C = 3,16 x 10^-5 mol/dm3
pOH = -log[OH-] = 4,5

pH = 14 - 4,5 = 9,5

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Answer

Proszę o wyprowadzenie tych wzorów bym wiedziała skąd np. z sin^2x powstaje tg^2/1+tg^2 i reszty

Zadanie Z10 Nie wychodzi mi wynik może ktoś wie jak je zrobić?

Zadanie Z10 Nie wychodzi mi wynik może ktoś wie jak je zrobić?

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