What is the $[O H^{-}]$ $[OH^-]$ for a solution at 25°C that has a $[H_{3} O^{+}]$ $[H_3O^+]$ of $5.36 \times 10^{- 5}$ $5.36 times 10^-5$ ?

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Answer 1 · 2017-06-22T04:26:03

$[H O^{-}] = 1.87 \times 10^{- 10} \cdot m o l \cdot L^{- 1}$ $[HO^-]=1.87xx10^-10*mol*L^-1$

Given.......

$2 H_{2} O (l) ⇌ H_{3} O^{+} + H O^{-}$ $2H_2O(l) rightleftharpoonsH_3O^+ + HO^-$

$K_w'=([H_3O^+][HO^-])/([H_2O(l)])$

Because $[H_2O]$ is so large, we remove it from the expression to give,,,,,,

$K_w=[H_3O^+][HO^-]=10^-14$ under standard conditions of temperature $(298*K)$ and pressure $(1*atm)$ ...

Thus $[HO^-]=10^-14/(5.36xx10^-5)=1.87xx10^-10*mol*L^-1$ .

Alternatively, we could take $log_10$ of both sides of $K_w=[H_3O^+][HO^-]=10^-14$ to give.........

$log_10K_w=log_10[H_3O^+]+log_10[HO^-]$

On rearrangement

$underbrace(-log_10[H_3O^+])_color(red)(pH)underbrace(-log_10[HO^-])_color(blue)(pOH)=underbrace(-log_10K_w)_(pK_w)$

And thus $color(red)(pH)+color(blue)(pOH)=pK_w=14$ .

And here, $pH=-log_10(5.36xx10^-5)=4.27$ , and clearly, $pOH=9.73$ .

$[HO^-]=10^(-9.73)=1.87xx10^-10*mol*L^-1$ , as required.......

I don't know about you, but I think the $log$ method is better.....

What is the [OH^-][OH−][OH^-] for a solution at 25°C that has a [H_3O^+][H3O+][H_3O^+] of 5.36 times 10^-55.36×10−55.36 times 10^-5?