What is $p H$ $pH$ for a solution whose $[H_{3} O^{+}] = 6.7 \times 10^{- 5} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=6.7xx10^-5molL^-1$ ?

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What is $p H$ $pH$ for a solution whose $[H_{3} O^{+}] = 6.7 \times 10^{- 5} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=6.7xx10^-5molL^-1$ ?

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Answer 1 · 2016-08-19T04:53:56

$p H = 4.17$ $pH=4.17$

Explanation:

$p H = {log}_{10} [H_{3} O^{+}]$ $pH=log_10[H_3O^+]$ (equivalently $p H = {log}_{10} [H^{+}]$ $pH=log_10[H^+]$ ).

Thus $p H = {log}_{10} (6.7 \times 10^{- 5})$ $pH=log_10(6.7xx10^-5)$ $=$ $=$ $- (- 4.17)$ $-(-4.17)$

Note that when we write ${log}_{a} b = c$ $log_ab=c$ , we are looking for the power to which we raise the base $a$ $a$ to get $b$ $b$ ; here $a^{c} = b$ $a^c=b$ . What are ${log}_{10} 100$ $log_(10)100$ , ${log}_{10} 1000000$ $log_(10)1000000$ , and ${log}_{10} 0.001$ $log_(10)0.001$ ?

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What is $p H$ $pH$ for a solution whose $[H_{3} O^{+}] = 6.7 \times 10^{- 5} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=6.7xx10^-5molL^-1$ ?

1 Answer

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What is pHpH for a solution whose [H3O+]=6.7×10−5⋅mol⋅L−1[H_3O^+]=6.7xx10^-5*mol*L^-1?

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What is $p H$ $pH$ for a solution whose $[H_{3} O^{+}] = 6.7 \times 10^{- 5} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=6.7xx10^-5molL^-1$ ?