I prepared $4\ \mathrm{M}$ solution of $\ce{NaOH}$, and the glass electrode I use cannot measure it the $\mathrm{pH}$ correctly. What are the other options do I have to exactly know the solution's $\mathrm{pH}$?
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$\begingroup$ This link explains the problem. all-about-ph.com/acid-and-alkaline.html The following link is for a high pH electrode but I couldn't find any specifications for it. :-( awedirect.co.uk/ph/ph-electrode/… $\endgroup$ – MaxW Feb 28 '17 at 17:39
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$\begingroup$ Do you really need to know pH or just that solution is 4.00 molar NaOH? Can you assume that the solution is "pure" NaOH? No NaCl, no KOH. I'm wondering about a sodium electrode, or conductivity. Carbonate from the air will be a problem. $\endgroup$ – MaxW Feb 28 '17 at 17:58
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$\begingroup$ Did you use tap water or distilled water? $\endgroup$ – Bob Mar 1 '17 at 0:45
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$\begingroup$ thanks ringo, I had calculated the same way, and I had made it in Miili q water, pH 5.7, and using AR grade NaOH. $\endgroup$ – angie Mar 3 '17 at 17:20
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For $\mathrm{pH}>12$ and $\ce{Li+}$ or $\ce{Na+}$ concentrations greater than $0.1\ \mathrm{M}$, glass electrodes experience alkali error. Because the concentration of $\ce{H3O+}$ ions in solution is so low, interference from alkali metal ions (like $\ce{Na+}$) becomes noticeable and causes the electrode to read lower than the true $\mathrm{pH}$ of the solution. $\ce{K+}$ typically cause less error than $\ce{Li+}$ or $\ce{Na+}$ due do its larger size.
Since sodium hydroxide is a strong base and dissociates completely into $\ce{Na+}$ and $\ce{OH-}$ ions, though, it's actually quite simple to just calculate the $\mathrm{pH}$.
$$K_\mathrm{w}=\ce{[H3O+][OH-]} \Rightarrow$$ $$ 1 \times 10^{-14} \ \mathrm{M^2}=\ce{[H3O+] \cdot 4\ \mathrm{M}} \Rightarrow $$ $$\ce{[H3O+]}=2.5 \times 10^{-15}\ \mathrm{M}$$ $$\mathrm{pH}=–\log(2.5 \times 10^{-15}\ \mathrm{M})=14.60$$
If you are worried about whether or not you truly have a $4\ \mathrm{M}$ solution of $\ce{NaOH}$, you should try titrating it with a known concentration of acid (in the past I always used potassium hydrogen phthalate). This process is called standardization in acid-base parlance.
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2$\begingroup$ I am pretty sure that is not 'exact'. If the electrode cannot measure it, there is a decent chance, that the final solution does not behave like an ideal solution. $\endgroup$ – Martin - マーチン♦ Feb 28 '17 at 9:34
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