At 25°C
PH+PoH=14
PH=14-PoH
PoH =-log[OH]
a)PH =14-(-log[.65×7.3×10^-10])
=14-9.323=4.677
b)PH=14-(-log[3.15×1.76×10^-5])=14-4.256=9.744
C)PH=14-(-log[.82×6.3×10^-5])=14-4.2868=9.7132
d)PH=14-(-log[.100×(2.5×10^-4+5.6×10^-8+3×10^-13)])
=14-4.60
=9.4
5. (8 points) Calculate the pH of each of the following aqueous solutions at 25 °C....
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