# pH 10 to pH9 NaOH dilution

I have a pH 10 NaOH solution. I need to bring down the pH to 9. So I used the $C_1V_1 = C_2V_2$ approach.

$C_1 = \mathrm{0.0001~mol/L}$

$V1$ is the necessary amount to produce 10 L of pH9

$C_2 = \mathrm{0.00001~mol/L}$

$V_2 = \mathrm{10~L}$

\begin{align} C_1V_1 &= C_2V_2\\ 0.0001~\mathrm{mol/L}\cdot V_1 &= 0.00001~\mathrm{mol/L}\cdot 10\mathrm{L}\\ V_1 &= \frac{(0.00001)(10)}{(0.0001)}\\ V_1 &= 1\mathrm{L}\\ \end{align}

I Applied this calculation in reality and used a pH probe and got a pH of 6.5. I think the drop is too sudden in this type of titration. What should I do?

• Double check the pH meter calibration. It is impossible to dilute NaOH and get an acidic solution. – MaxW Feb 3 '16 at 22:13
• And who says water is neutral? There's always some CO2 in it, unless it wouldn't have any contact with air. – Mithoron Feb 3 '16 at 22:28
• When you dilute anything down to homeopathic concentrations like 0.00001, expect all kinds of weird effects. Any impurity ($\ce{CO2}$ from air, like Mithoron said) could suddenly become more abundant than your compound. – Ivan Neretin Feb 4 '16 at 5:49
• Even with distilled water? What would be a new approach I should take? My problem is I need to make a pH10 solution which I need to creata pH9, pH8 and pH7 from the pH10 solution . – user25260 Feb 4 '16 at 13:08
• Yes, even with distilled water. The distilled water condenses out of the air, and voila! Carbon dioxide! – Oscar Lanzi May 25 '16 at 2:16

Generally, you don't make pH of $7\pm2$ by dilution alone. The required dilution is so great that it would leave very little of your compound, and then any tiny impurity would affect the result greatly. Think of a droplet on the pH meter probe left from the previous measurement. Distilled water does not help much. All the same, it contains $\ce{CO2}$ from the air, so its pH is usually about 5.5, not 7 as one might expect.