Kw = [H+][OH-] = 1*10-14
pH=−lg[H+]=−log10(4∗10−11)=10.4pH = -lg[H^+]=-log_{10}(4*10^{-11})=10.4pH=−lg[H+]=−log10(4∗10−11)=10.4
pOH=14−pH=3.6pOH = 14-pH = 3.6pOH=14−pH=3.6
[OH−]=10−pOH=0.00025=2.5∗10−4[OH^-]=10^{-pOH}=0.00025=2.5*10^{-4}[OH−]=10−pOH=0.00025=2.5∗10−4
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments