Open Question: Finding pOH in a Chemical Reaction?
23:54 Publicado por Flechado16
Calculate the pOH of a 0.0827 M aqueous sodium cyanide solution at 25.0 °C. Kb for CN- is 4.49*10^(-10).
Do I use this equation?
NaCN(aq) + H2O(l) ? [Na+](aq) + [OH-](aq) + [H+](aq) + [CN-](aq)
? Kb/Kw = [Na+][CN-]/[NaCN] = x²/(0.0827 - x) = 4.49*10^4
---? x = 0.0826998 M = [Na+] = [CN-]
But then how would I get a pOH out of that? adding NaCN doesn't directly affect the concentration... or am I supposed to look up the pH of [CN-] on a table or something? I'm sorry, I really have no idea how to solve this.
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