# Measuring pH of pasty solutions/mixtures

I'm trying to measure the pH of a very viscous paste mixture and have been told to dilute the sample for more accurate results, because the viscosity of the sample may prevent the sensor to detect the real pH of the sample.

Does this sound right to you? Wouldn't diluting the sample alter the pH?

Thank you

• Yes, this does sound right. And yes, of course it would change the pH. The advice was probably given in the hope that you'd be able to account for that change. This, BTW, is not always easy and not even always possible. Dec 22, 2016 at 14:34
• – user7951
Dec 22, 2016 at 14:40
• Do you know the acids/bases in your system? Are they weak acids/bases?
– NMJD
Dec 22, 2016 at 15:14

Diluting the sample does alter the pH, but it does so in a predictable way.

When we are measuring pH, we are measuring the concentration of hydrogen ions (or hydornium ions, or whatever ) via the negative logarithm:

$$\mathrm{pH} = -\log_{10}\left([\ce{H+}]\right)$$

Diluting a solution has a predictable effect on concentration, since the total amount (in moles) of $\ce{H+}$ will not change by the addition of more solvent. This relationship is summarized by the formula (where $c$ represents concentration, $v$ represents volume, and $1$ and $2$ represent the solution before and after dilution, respectively:

$$c_1v_1 =c_2v_2$$

Let's take an example. Suppose you have 1.000 mL of a viscous acidic aqueous solution. The viscosity of the solution will impede easy measurement of pH because of the kinetics of mass transport. If you dilute this solution with water to a total volume of 10.00 mL, you now have a less viscous solution with the same number of moles of $\ce{H+}$, because sources of $\ce{H+}$ were neither added nor removed.

If this new diluted solution has $\mathrm{pH = 3.03}$, then $$[\ce{H+}] = 10^{\mathrm{-pH}}=10^{-3.03} = 9.33\times 10^{-4}\ \mathrm{M}$$.

The concentration of the original solution can now be determined:

$$c_1 = \dfrac{c_2 v_2}{v_1} = \dfrac{\mathrm{9.33\times 19^{-4}\ M)(10.00\ mL)}}{\mathrm{1.000\ mL}}=9.33\times 10^{-3}\ \mathrm{M}$$

The pH of the original solution is:

$$\mathrm{pH = -\log_{10}(9.33\times 10^{-3}) = 2.03}$$

Note, that since I chose to dilute by a factor of 10, the pH of the dilute solution is 1 pH unit higher than the original solution.

This approach works for solutions of strong acids and strong bases and will work over much of the pH range, but may get a little awkward around neutral. The approach breaks down for weak acids and bases since changing the concentration changes the position of equilibrium. For weak acids, diluting likely results in additional ionization events. However, even without knowing the acid/base and its equilibrium constant, you could perform multiple dilutions and do some curve-fitting and get back to the original pH.