1
$\begingroup$

The pH of a solution is 7.00. To this solution sufficient base is added to increase the pH to 12.0. the increase $\ce{OH-}$ ion concentration is:

(A) $5$ times
(B) $1000$ times
(C) $10^5$ times
(D) $4$ times

The answer is (C) $10^5$ times.

Now, the temperature hasn't been mentioned ergo we can't assume that it's a neutral soultion, right? So, I tried doing it this way:

\begin{array}{llll} &\mathrm{pH} &= 7 &\qquad &\mathrm{pH} &= 12 \\ &[\ce{H+}] &= 10^{-7} &\qquad &[\ce{H+}] &= 10^{-12} \\ & & &[\ce{OH-}]~\text{added} \\ & & &= 10^{-7} - 10^{-12} \\ & & &= 10^{-7} (1 - 10^{-5}) \\ & & &\approx 10^{-7} \end{array}

$[\ce{OH-}]$ added is the increase in its concentration, or so I think. I don't know what I'm doing wrong. Please, help.

Also, can you please point out the loopholes I might have in my concept if it's obvious from my doubt, so I can look up and take care of it?

$\endgroup$
4
$\begingroup$

The problem is that you are trying to subtract the values, whereas all you need to do is find an increase in ratio $\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1}$ knowing that $\mathrm{pOH} = - \log{[\ce{OH-}]}$ and $\mathrm{pOH} = \mathrm{p}K_\mathrm{w} - \mathrm{pH}$:

$$\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1} = \frac{10^{-\mathrm{pOH_2}}}{10^{-\mathrm{pOH_1}}} = 10^{\mathrm{pOH_1} - \mathrm{pOH_2}} = 10^{\mathrm{p}K_\mathrm{w} - \mathrm{pH_1} - \mathrm{p}K_\mathrm{w} + \mathrm{pH_2}} = 10^{\mathrm{pH_2} - \mathrm{pH_1}} = 10^{12 - 7} = 10^5$$

$\endgroup$
  • 1
    $\begingroup$ pOH=14−pH is true only for 25°C. So,unless it's specified otherwise, are we to assume that the dissociation is taking place at 25C? $\endgroup$ – Jamil Ahmed Oct 15 '17 at 9:29
  • 1
    $\begingroup$ @JamilAhmed Yep, you are right, but considering the problem is not mentioning temperature at all, I assumed it's safe not to overthink and go for the easiest solution:) $\endgroup$ – andselisk Oct 15 '17 at 9:31
  • 2
    $\begingroup$ @JamilAhmed It's just basic MarkDown and MathJax, if you feel like you want to know a bit more, I'd suggest to read MathJax basic tutorial and quick reference and How can I format math/chemistry expressions here?. $\endgroup$ – andselisk Oct 15 '17 at 10:04
  • 2
    $\begingroup$ @JamilAhmed You only need to know that $\mathrm{pOH_1} + \mathrm{pH_1} = \mathrm{pOH_2} + \mathrm{pH_2}$, i.e. that the constant (the logarithm of the self-ionization constant) that is 14 at 25°C is the same before and after adding $\ce{OH-}$. $\endgroup$ – wythagoras Oct 15 '17 at 11:47
  • 1
    $\begingroup$ @wythagoras Good point, I changed $\mathrm{pOH} = 14 - \mathrm{pH}$ to $\mathrm{pOH} = \mathrm{p}K_\mathrm{w} - \mathrm{pH}$ in order to address the doubt about the temperature. Thank you! $\endgroup$ – andselisk Oct 15 '17 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.