calculate the ph of 0.05M of NaOH?
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NaOH = Na+ + OH-
so
0.05 M NaOH ---> 0.05 M OH- (1 to 1 ratio)
then
p(OH) = -log[OH-] = -log(0.05) = 1.3
then
pH + p(OH) = 14.0 1.0x10^-14)
and
pH = 14.0 - p(OH) = 14.0 - 1.3 = 12.7
so
0.05 M NaOH ---> 0.05 M OH- (1 to 1 ratio)
then
p(OH) = -log[OH-] = -log(0.05) = 1.3
then
pH + p(OH) = 14.0 1.0x10^-14)
and
pH = 14.0 - p(OH) = 14.0 - 1.3 = 12.7
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