What is the pH of a 0.0035M KOH solution?
A)2.46
B)5.65
C)8.35
D)11.54
E)None of these
*the answer is D)11.54 but I don't know how to get it.
KOH(aq) -------> K+(aq) + (OH-)(aq)
Now, Sicne KOH is a strong base, therefore it dissociates completely
Hence [OH-] = 0.0035 M
Now, [H+]*[OH-] = 1*10-14
or, [H+] = 2.857*10-12
Thus, pH = -log[H+] = 11.544
[OH-1] = .0035 M
pH = 14 - log (OH-1 )
pH = 14 - log (.0035) = 11.54
D)11.54
[OH-] = .0035 M
pOH = - log (.0035) = 2.455
pH = 14.00 - pOH = 14.00 - 2.455 = 11.54
What is the pH of a 0.0035M KOH solution? A)2.46 B)5.65 C)8.35 D)11.54 E)None of these...
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