What is the pH of an aqueous solution if the [ $H^{+}$ $H^+$ ]=0.000001?

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Answer 1 · 2016-10-03T17:07:41

$p H = 6$ $pH=6$

$p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_10[H_3O^+]$

But if $[H^{+}]$ $[H^+]$ or $[H_{3} O^{+}]$ $[H_3O^+]$ $=$ $=$ $0.000001 \cdot m o l \cdot L^{- 1}$ $0.000001*mol*L^-1$ ;

then $p H$ $pH$ $=$ $=$ $- {log}_{10} (0.000001)$ $-log_10(0.000001)$ $=$ $=$ $- {log}_{10} (10^{- 6})$ $-log_10(10^-6)$ .

Now, by definition, ${log}_{10} (10^{- 6})$ $log_10(10^-6)$ $=$ $=$ $- 6$ $-6$ .

Thus $p H = - (- 6)$ $pH=-(-6)$ $=$ $=$ $6$ $6$

This is slightly acid, and $[H_{3} O^{+}]$ $[H_3O^+]$ is $10 \times [H O^{-}]$ $10xx[HO^-]$ .

If $[H_{3} O^{+}]$ $[H_3O^+]$ $=$ $=$ $0.001 \cdot m o l \cdot L^{- 1}$ $0.001*mol*L^-1$ , what is $p H$ $pH$ ?

What is the pH of an aqueous solution if the [H+H^+]=0.000001?