The pOH of beer is 10.0. What is the concentr | vedclass

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(B, D) Given: $pOH = 10.0$. We know that $[OH^-] = 10^{-pOH} = 10^{-10} \ M$. Using the ionic product of water,$K_w = [H^+][OH^-] = 10^{-14}$. Therefo

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$10^{-4}$

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(B, D) Given: $pOH = 10.0$. We know that $[OH^-] = 10^{-pOH} = 10^{-10} \ M$. Using the ionic product of water,$K_w = [H^+][OH^-] = 10^{-14}$. Therefore,$[H^+] = \frac{K_w}{[OH^-]} = \frac{K_w}{10^{-10}}$. Also,$[H^+] = 10^{-pH}$. Since $pH + pOH = 14$,$pH = 14 - 10 = 4$. Thus,$[H^+] = 10^{-4} \ M$. Both options $(b)$ and $(d)$ are correct.

The $pOH$ of beer is $10.0$. What is the concentration of hydrogen ions $[H^+]$?

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