# Find increase in OH- ion concentration after base is added

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?

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 $\mathrm{pOH} = \mathrm{p}K_\mathrm{w} - \mathrm{pH}$:
$$\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1} = \frac{10^{-\mathrm{pOH_2}}}{10^{-\mathrm{pOH_1}}} = 10^{\mathrm{pOH_1} - \mathrm{pOH_2}} = 10^{\mathrm{p}K_\mathrm{w} - \mathrm{pH_1} - \mathrm{p}K_\mathrm{w} + \mathrm{pH_2}} = 10^{\mathrm{pH_2} - \mathrm{pH_1}} = 10^{12 - 7} = 10^5$$
• @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 $\ce{OH-}$. – wythagoras Oct 15 '17 at 11:47
• @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! – andselisk Oct 15 '17 at 13:33