How do you determine pH from molarity?

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How do you determine pH from molarity?

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Answer 1 · 2014-03-06T02:48:37

Molarity is also known as the concentration. The concentration can be converted to $pH$ $"pH"$ .

Take a sample of nitric acid ${HNO}_{3}$ $"HNO"_3$ with a molarity of $0.0032 M$ $"0.0032 M"$ .

The molarity becomes the $[H^{+}]$ $["H"^+]$ concentration of $3.2 \times 10^{- 3}$ $3.2 xx 10^(-3)$ .

We know this is the $[H^{+}]$ $["H"^+]$ concentration because ${HNO}_{3}$ $"HNO"_3$ is a strong acid, since the acid will completely dissociate into $H^{+}$ $"H"^+$ and ${NO}_{3}^{-}$ $"NO"_3^-$ .

To find the $pH$ $"pH"$ calculate $- log [H^{+}]$ $- log["H"^+]$

$- log [3.2 \times 10^{- 3}] = 2.49 = pH$ $-log[3.2xx 10^(-3)] = 2.49 = "pH"$

To find the $pOH$ $"pOH"$ take the $pH$ $"pH"$ value and subtract it from $14$ $14$ .

$14 - 2.49 = 11.51 = pOH$ $14 - 2.49 = 11.51 = "pOH"$

To convert $pOH$ $"pOH"$ to $[{OH}^{-}]$ $["OH"^-]$ calculate $[{OH}^{-}] = 10^{- pOH}$ $["OH"^-]=10^(-"pOH")$

$10^{- 11.51} = 3.09 \times 10^{- 12} = [{OH}^{-}]$ $10^(-11.51)= 3.09 xx 10^(-12)=["OH"^-]$

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How do you determine pH from molarity?

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