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Sagot :
Steps:-
- First we calculate pH then pOH then [OH-]
[tex]\\ \tt\rightarrowtail pH=-log[H^+][/tex]
[tex]\\ \tt\rightarrowtail pH=-log[9.1\times 10^{-4}][/tex]
[tex]\\ \tt\rightarrowtail pH=-log9.1-log10^{-4})[/tex]
[tex]\\ \tt\rightarrowtail pH=0.95+4[/tex]
[tex]\\ \tt\rightarrowtail pH=4.95[/tex]
Now
[tex]\\ \tt\rightarrowtail pH+pOH=14[/tex]
[tex]\\ \tt\rightarrowtail pOH=14-4.95[/tex]
[tex]\\ \tt\rightarrowtail pOH=9.05[/tex]
So
[tex]\\ \tt\rightarrowtail -log[OH^-]=9.05[/tex]
[tex]\\ \tt\rightarrowtail log[OH^-]=-9.05[/tex]
[tex]\\ \tt\rightarrowtail OH^-=10^{-9.05}[/tex]
[tex]\\ \tt\rightarrowtail OH^-=8.91\times 10^{-4}[/tex]
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