Respuesta :
The answer is [tex]\underline{12.954}\end{aligned}$[/tex]
[tex]\begin{aligned}&\mathrm{pH}=\underline{1.046} \\&{\left[\mathrm{OH}^{-}\right]=\underline{1.11 \times 10^{-13} \mathrm{M}}} \\&\mathrm{pOH}=\underline{12.954}\end{aligned}[/tex]
Given:
[tex]\left[\mathrm{H}^{+}\right]=0.090 \mathrm{M}=9 \times 10^{-2} \mathrm{M} ; \mathrm{T}\\=25^{\circ} \mathrm{C}$\mathrm{As}, \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$\Rightarrow \mathbf{p H}=-\log \left(9 \times 10^{-2}\right)=\underline{1.046}$[/tex]
The equation for the self-ionization constant of water is
[tex]$\mathrm{Kw}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]$and, $\mathrm{pKw}=\mathrm{pH}+\mathrm{pOH}$[/tex]
Since at room temperature,
[tex]$25^{\circ} \mathrm{C}: \mathrm{Kw}=1.0 \times 10^{-14}, \mathrm{pKw}=14$[/tex]
[tex]\therefore \mathrm{Kw}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-14}$\\\Rightarrow\left[\mathrm{OH}^{-}\right]=\left(1.0 \times 10^{-14}\right) \div\left[\mathrm{H}^{+}\right]\\=\left(1.0 \times 10^{-14}\right) \div\left[9 \times 10^{-2}\right]\\=0.111 \times 10^{-12}=1.11$\times 10^{-13} \mathrm{M}$[/tex]
And
[tex]$\begin{aligned}&\mathrm{pH}+\mathrm{pOH}=\mathrm{pKw}=14 \\&\Rightarrow \mathrm{pOH}=14-\mathrm{pH}=14-1.046=\underline{12.954}\end{aligned}$[/tex]
What is Poh and PH linked formula?
- Take the negative log of the hydronium ion concentration and use that value to compute pH.
- Simply subtraction the pH from 14 yields the pOH value.
- The negative log of the concentration of hydroxide ions should be used to get the pOH.
- Simply subtraction 14 from pOH yields the pH.
So the more about Poh and PH linked formula.
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