# Is there a concise way to calculate pH of a solution after being diluted with water?

I do not have a chemistry degree or even much knowledge in the subject at all. I'm actually an auto detailer and this is why I have an interest in pH levels. All of our chemicals obviously have pH levels associated with them. I am trying to figure out a way to calculate the new pH of a solution when diluted with water. I have tried to do some research online about it but it brings in things like molarity, of which I know nothing. I'll give an example of what I need:

This chemical has a pH of "12-13" according to the SDS sheet. So let's say 13 for simplicity.

We use this chemical in two different dilutions: 10 parts water to 1 part chemical, and 4 parts water to 1 part chemical.

How would I calculate the pH level of the new solution?

• Simplest way is to just measure it. Unless you know what compounds are in there, you could only make a guess (probably wrong one). If everything about compounds was simple there wouldn't be much need for chemistry as a separate science discipline. Nov 27, 2022 at 13:58
• It would help alot if could list the chemicals in your pH 13 solution. But, just for example, diluting a pH 13 solution of caustic soda (lye) 1 to 10 will still be a very strongly basic pH 12. This is because pH numbers are logarithmic. Nov 27, 2022 at 22:15

The calculation will depend on whether the chemical of interest predominantly behaves as an acid or a base, whether it's strong or weak, and whether it's monoprotic or polyprotic.

Concentrated pure aqueous solutions of a chemical with a pH below 7 are typically acids, while concentrated pure aqueous solutions of a chemical with a pH above 7 are typically bases.

Assuming the chemical "B" is a strong monoprotic base and water autoionization is negligible, the moles of hydroxide ions $$\ce{OH-}$$ of a solution with volume $$V_B$$, before dilution are:

$$n_{\ce{OH-}}=10^{\pu{-pOH_o}}\;V_B$$

After diluting by adding a certain volume $$V_w$$ of pure water:

$$C_{\ce{OH-}}=10^{\pu{-pOH_o}}\frac{V_B}{V_B+V_w}$$

Taking the negative base 10 logarithm on both sides:

$$\pu{-log}\;C_{\ce{OH-}}=\pu{-log}\;(10^{\pu{-pOH_o}})-\pu{log}\;\left(\frac{V_B}{V_B+V_w}\right)$$

Which is equivalent to:

$$\pu{pOH}=\pu{pOH_o}-\pu{log}\;\left(\frac{V_B}{V_B+V_w}\right)$$

Or in terms of pH:

$$14-\pu{pH}=14-\pu{pH_o}-\pu{log}\;\left(\frac{V_B}{V_B+V_w}\right)$$

Solving for pH:

$$\pmb{\pu{pH}=\pu{pH_o}+\pu{log}\;\left(\frac{V_B}{V_B+V_w}\right)\quad\text{(for SMB)}}$$

For a strong monoprotic acid, the derivation is analogous, except no conversion between $$\pu{pH}$$ and $$\pu{pOH}$$ is necessary, so the resulting expression is:

$$\pmb{\pu{pH}=\pu{pH_o}+\pu{log}\;\left(\frac{V_A+V_w}{V_A}\right)\quad\text{(for SMA)}}$$

Where in both cases, pH represents the final value (after dilution), while $$\pu{pH_o}$$ represents the initial value (before dilution).

For either case, adding a very high volume of water will make the pH of the solution asymptotically approach the value of 7, but it will never reach it.

Some examples of monoprotic strong bases are sodium hydroxide ($$\ce{NaOH}$$) and potassium hydroxide ($$\ce{KOH}$$).

Some examples of monoprotic strong acids are hydrochloric acid ($$\ce{HCl}$$) and hydrobromic acid ($$\ce{HBr}$$).

For a weak monoprotic acid, the calculation is as follows:

The initial concentration of the acid solution can be set in terms of $$x$$:

$$C_{Ao}=\frac{x^2}{K_a}+x=x\left(\frac{x}{K_a}+1\right)$$

Which means the total moles of acid before dilution is:

$$n_{Ao}=C_{Ao}V_A=x\left(\frac{x}{K_a}+1\right)V_A$$

At equilibrium, before dilution, the equilibrium constant expression in terms of moles is:

$$K_{n_o}=K_a\;V_A=\frac{y^2}{n_{Ao}-y}$$

At equilibrium, after dilution, the equilibrium constant expression in terms of moles is:

$$K_{n}=K_a\;(V_A+V_w)=\frac{(y+z)^2}{n_{Ao}-(y+z)}$$

Substituting $$n_{Ao}$$:

$$K_a\;(V_A+V_w)=\frac{(y+z)^2}{x\left(\frac{x}{K_a}+1\right)V_A-(y+z)}$$

Where:

$$x=10^{-\pu{pH_o}}$$

$$y+z=10^{-\pu{pH}}\;(V_A+V_w)$$

Substituting above and simplifying:

$$K_a=\frac{10^{-\pu{pH}}}{10^{\pu{pH-pH_o}}\left(\frac{10^{-\pu{pH_o}}}{K_a}+1\right)\frac{V_A}{V_A+V_w}-1}$$

Solving for pH:

$$\pu{pH}=\pu{log}\left(\frac{K_a+\sqrt{K_a^2+4\times10^{-\pu{pH_o}}\;(10^{-\pu{pH_o}}+K_a)\;\frac{V_A}{V_A+V_w}}}{2\times10^{-\pu{pH_o}}\;(10^{-\pu{pH_o}}+K_a)\;\frac{V_A}{V_A+V_w}}\right)$$

Conversely, for a weak monoprotic base:

$$K_b=\frac{10^{\pu{pH}-14}}{10^{\pu{pH_o-pH}}\left(\frac{10^{\pu{pH_o}-14}}{K_b}+1\right)\frac{V_B}{V_B+V_w}-1}$$

Solving for pH:

$$\pu{pH}=14-\pu{log}\left(\frac{K_b+\sqrt{K_b^2+4\times10^{\pu{pH_o}-14}\;(10^{\pu{pH_o}-14}+K_b)\;\frac{V_B}{V_B+V_w}}}{2\times10^{\pu{pH_o}-14}\;(10^{\pu{pH_o}-14}+K_b)\;\frac{V_B}{V_B+V_w}}\right)$$

For simplicity, we can define:

$$\beta_A=2\times10^{-\pu{pH_o}}\;(10^{-\pu{pH_o}}+K_a)\;\frac{V_A}{V_A+V_w}$$

$$\beta_B=2\times10^{\pu{pH_o}-14}\;(10^{\pu{pH_o}-14}+K_b)\;\frac{V_B}{V_B+V_w}$$

So the final expressions become:

$$\pmb{\pu{pH}=\pu{log}\left(\frac{K_a+\sqrt{K_a^2+2\beta_A}}{\beta_A}\right)\quad\text{(for WMA)}}$$

$$\pmb{\pu{pH}=14-\pu{log}\left(\frac{K_b+\sqrt{K_b^2+2\beta_B}}{\beta_B}\right)\quad\text{(for WMB)}}$$

An example of a weak monoprotic acid is acetic acid ($$\ce{CH3COOH}$$).

An example of a weak monoprotic base is ammonia ($$\ce{NH3}$$).

The calculation for polyprotic acids and bases is significantly more complex.

As a final note, all volumes need to be in liters (L), and the dissociation constants $$K_a$$ and $$K_b$$ are dimensionless.

How would I calculate the pH level of the new solution?

There is no general formula to calculate this. It depends on what components are present in the solution.

I am trying to figure out a way to calculate the new pH of a solution when diluted with water. [...] We use this chemical in two different dilutions: 10 parts water to 1 part chemical, and 4 parts water to 1 part chemical.

If the solution is buffered, the pH changes little when diluting by these amounts. When the solution is not buffered, the pH moves a bit in the direction of neutral (i.e. pH 13 is now pH 12, and pH 2 is now pH 3).

The easiest estimate is to assume the pH did not change much, and have the same precautions as for the undiluted stuff ("chemical").

From this reference we find:

pH + pOH = 14

Log of [OH] = -1. [OH] is 10$$^-1$$ or 0.10 M.

Assuming the pH of the diluent is around 7, very little of the hydroxide will be neutralized. The solution is only diluted.

A pH 13 solution of NaOH diluted 1 to 10 will be 0.010 M. Log of [OH] = -2. Working the above formula, pH still be a very strongly basic pH 12.

Similarly, a pH 13 solution that is diluted 1 to 4 will be 0.025 M. Log of [OH] = -1.6, yielding a pH of 12.4.

From a safety standpoint, both diluted solutions remain highly basic.