Determine the pH of a 0.09 M KOH solution. Your answer must be within ± 0.1%?

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Determine the pH of a 0.09 M KOH solution. Your answer must be within ± 0.1%?

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Answer 1 · 2018-05-04T04:14:26

$pH = 12.95$ $"pH" = 12.95$

Explanation:

$KOH$ $"KOH"$ is a strong base, meaning that it dissociates completely. Therefore, the (balanced) equation is:
$KOH (aq) \to K^{+} (aq) + {OH}^{-}$ $"KOH (aq)" -> "K"^+ "(aq)" + "OH"^-$ $(a q)$ $(aq)"$

This balanced equation tells us that every mole of $KOH$ $"KOH"$ produces one mole of ${OH}^{-}$ $"OH"^-$ .
Therefore, we know that $[KOH] = [{OH}^{-}] = 0.09 M$ $["KOH"] = ["OH"^-] = 0.09 " M"$

Now, to find the $pOH$ $"pOH"$ , we use the formula:
$pOH = - log [{OH}^{-}]$ $"pOH" = -log["OH"^-]$

So:
$pOH = - log [0.09 M] \approx 1.05$ $"pOH" = -log[0.09 " M"] ~~ 1.05$

However, the question wants the $pH$ $"pH"$ , not the $pOH$ $"pOH"$ . To find this, we use:
$pH = 14 - pOH$ $"pH" = 14 - "pOH"$

So:
$pH = 14 - 1.05 = 12.95$ $"pH" = 14 - 1.05 = 12.95$

The exact solution to $8$ $8$ decimal places is $12.95424251$ $12.95424251$ . This is within $0.1 %$ $0.1%$ :

$∣ ∣ ∣ \frac{12.95 - 12.95424251}{12.95424251} ∣ ∣ ∣ \times 100 % \approx 0.03 %$ $|(12.95 - 12.95424251)/(12.95424251)| xx 100% ~~ 0.03%$

Hope this helps!

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Determine the pH of a 0.09 M KOH solution. Your answer must be within ± 0.1%?

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