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The H⁺ concentration in an aqueous solution at 25 °C is 9.1 × 10⁻⁴. What is [OH⁻]?

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]