What is $[H O^{-}]$ $[HO^-]$ in an aqueous solution for which $[H_{3} O^{+}] = 3.2 \times 10^{- 3} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=3.2xx10^-3molL^-1$ ?

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What is $[H O^{-}]$ $[HO^-]$ in an aqueous solution for which $[H_{3} O^{+}] = 3.2 \times 10^{- 3} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=3.2xx10^-3molL^-1$ ?

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Answer 1 · 2017-07-30T19:28:48

$[H O^{-}] = 3.16 \times 10^{- 12} \cdot m o l \cdot L^{- 1}$ $[HO^-]=3.16xx10^-12*mol*L^-1$

Explanation:

We know that water undergoes autoprotolysis according to the following equation......

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

WE know by precise measurement at $298 \cdot K$ $298*K$ under standard conditions, the ion product......

$K_{w} = [H_{3} O^{+}] [H O^{-}] = 10^{- 14}$ $K_w=[H_3O^+][HO^-]=10^-14$ .

Now this is an equation, that I may divide, subtract, multiply, PROVIDED that I does it to both sides. One thing I can do is to take ${log}_{10}$ $log_10$ of BOTH SIDES........

${log}_{10} {K_{w}} = {log}_{10} {[H_{3} O^{+}] [H O^{-}]} = {log}_{10} (10^{- 14})$ $log_10{K_w}=log_10{[H_3O^+][HO^-]}=log_10(10^-14)$

And thus ${log}_{10} [H_{3} O^{+}] + {log}_{10} [H O -] = - 14$ $log_10[H_3O^+]+log_10[HO-]=-14$ , and on rearrangement.....

$14=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)$

And so (FINALLY) we gets our working relationship, which you should commit to memory.....

$pH+pOH=14$

We have $[H_3O^+]=3.2xx10^-3*mol*L^-1$ . $pH=-log_10(3.2xx10^-3)=2.50$ . And thus $pOH=11.50$ .

And now we take antilogs..... $[HO^-]=10^(-11.50)*mol*L^-1$

$=3.16xx10^-12*mol*L^-1$ as required............

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What is $[H O^{-}]$ $[HO^-]$ in an aqueous solution for which $[H_{3} O^{+}] = 3.2 \times 10^{- 3} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=3.2xx10^-3molL^-1$ ?

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What is [HO^-][HO−][HO^-] in an aqueous solution for which [H_3O^+]=3.2xx10^-3*mol*L^-1[H3O+]=3.2×10−3⋅mol⋅L−1[H_3O^+]=3.2xx10^-3*mol*L^-1?

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What is $[H O^{-}]$ $[HO^-]$ in an aqueous solution for which $[H_{3} O^{+}] = 3.2 \times 10^{- 3} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=3.2xx10^-3molL^-1$ ?