What is the pH of $1 * 10^-4$ $M$ $NaOH$ ?

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Answer 1 · 2016-05-28T07:03:44

$pH = 10$

In water the following equilibrium operates:

$2H_2O(l) rightleftharpoonsH_3O^+ + HO^-$

As for any equilibrium, we can write the equilibrium expression:

$([H_3O^+][HO^-])/[[H_2O]]$ $=$ $K'_w$

Since $[H_2O]$ is huge, we can simplify this expression:

$[H_3O^+][HO^-]$ $=$ $K_w$

At $298K$ , $[H_3O^+][HO^-]$ $=$ $K_w=10^(-14)$

Taking $log_10$ of both sides:

$log_10[H_3O^+] + log_10[HO^-]$ $=$ $log_10(10^-14)$ $=$ $-14$

Now we can define $pH$ $=$ $-log_10[H_3O^+]$ , and

$pOH$ $=$ $-log_10[HO^-]$ .

And thus $pOH +pH=14$ $"(finally!)"$

Thus $pOH$ of $1xx10^-4mol*L^-1" NaOH"$ $=$ $-log_10(1xx10^-4)=4$

Since $pOH +pH=14$ , $pH=10$ , as required.

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