PART 1
Given : [ H + ] = 0.60 M
We have relation, [ H + ] [ OH - ] = K w = 1.00 10 -14
[ OH - ] = 1.00 10 -14 / [ H + ]
[ OH - ] = 1.00 10 -14 / 0.60
[ OH - ] = 1.67 10 -14
We have , pH = - log [ H + ]
pH = - log 0.60 = 0.22
PART 2
Given : [ OH - ] = 9.9 10 -12 M
We have relation, [ H + ] [ OH - ] = K w = 1.00 10 -14
[ H + ] = 1.00 10 -14 / [ OH - ]
[ H + ] = 1.00 10 -14 / 9.9 10 -12
[ H + ] = 1.01 10 -03 M
We have , pH = - log [ H + ]
pH = - log 1.01 10 -03= 2.30
PART 3
Given : pH = 1.83
We have , pH = - log [ H + ]
[ H + ] = 10 - pH
[ H + ] = 10 - 1.83 = 0.01479 M
We have, [ OH - ] = 1.00 10 -14 / [ H + ]
[ OH - ] = 1.00 10 -14 / 0.01479
[ OH - ] = 6.8 10 -13 M
calculate the [OH-] and the pH of a solution with an CH] =0.60 M at 25°C....
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