5.2*10^-4 oH- solution
Answers
Answer:
For pure water at 25∘C, the concentration of hydronium ions, H3O+, is equal to the concentration of hydroxide ions, OH−.
More specifically, water undergoes a self-ionization reaction that results in the formation of equal concentrations of hydronium and hydroxide anions.
2H2O(l]⇌H3O+(aq]+OH−(aq]
At room temperature, the self-ionization constant of water is equal to
KW=[H3O+]⋅[OH−]=10−14
This means that neutral water at this temperature will have
[H3O+]=[OH−]=10−7M
As you know, pH and pOH are defined as
pH=−log([H3O+])
pOH=−log([OH−])
and have the following relationship
pH+pOH=14
In your case, the concentration of hydroxide ions is bigger than 10−7M, which tells you that you're dealing with a basic solution and that you can expect the pH of the water to be higher than 7.
A pH equal to 7 is characteristic of a neutral aqueous solution at room temperature.
So, you can use the given concentration of hydroxide ions to determine the pOH of the solution first
pOH=−log(4.62⋅10−4)=3.34
This means that the solution's pH will be