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Answer 1 · 2017-08-22T07:33:20

$[H O^{-}] = 1.78 \times 10^{- 11} \cdot m o l \cdot L^{- 1}$ $[HO^-]=1.78xx10^-11*mol*L^-1$

In aqueous solution, water undergoes the famous autoprotolysis reaction:

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

And under standard condition of $298 \cdot K$ $298*K$ , and near $1 \cdot a t m$ $1*atm$ ...

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

Taking ${log}_{10}$ $"log_10$ of BOTH sides........

${log}_{10} K_{w} = {log}_{10} [H O^{-}] + {log}_{10} [H_{3} O^{+}]$ $log_10K_w=log_10[HO^-]+log_10[H_3O^+]$ .....on rearrangment...

$underbrace(-log_10[HO^-])_(pH)underbrace(-log_10[H_3O^+])_(pH)=underbrace(-log_10(10^-14))_14$

And thus our working relationship.......which you will have to commit to memory:

$pH+pOH=14$

We have $pH=3.25$ , thus $pOH=10.75$ , and .........

$[HO^-]=10^(-10.75)=1.78xx10^-11*mol*L^-1$ ......

For the use of $pH$ in buffer equations, see [here.](https://socratic.org/questions/how-do-buffers-maintain-ph)