the pH of a solution is 10 and other is 12 when equal volumes of these low are mixed the pH of the resulting solution is?

"equal volumes of these low" WHAT?

If the acids are strong acids we can take 100 mL of each.
For pH = 10 we have 100 x 1E-10 = 1E-8 millimols.
Add 100 mL pH = 12 and we have 100 x 1E-10 millimols.
Add 1E-8 mmols and 1E-10 mmols = 1.01E-8 millimols and that is in 200 mL so
(H^+) = 1E-8 mmols/200 mL = 5.05E-11 and pH = ?. About 10.3
You can use 50 mL each or 1000 mL each ( or any equal numbers) and the answer will be the same.

To find the pH of the resulting solution when equal volumes of two solutions with different pH values are mixed, we can use the concept of dilution and the fact that pH is a logarithmic scale.

First, we need to understand that pH is a measure of the acidity or alkalinity of a solution. It is defined as the negative logarithm (base 10) of the hydrogen ion concentration ([H+]) in the solution. Mathematically, the pH formula is written as pH = -log[H+].

In this case, we have two solutions with pH values of 10 and 12. Taking the negative logarithm of the hydrogen ion concentration, we can determine that the [H+] for a pH of 10 is 10^-10 moles per liter, and for a pH of 12, it is 10^-12 moles per liter.

When equal volumes of these solutions are mixed together, the total volume doubles, but the total number of moles of hydrogen ions remains the same. Therefore, the hydrogen ion concentration in the resulting solution will be the average of the two initial concentrations.

Let's calculate the average [H+] concentration:

[H+] average = (10^-10 + 10^-12) / 2

Now, we can convert the average [H+] concentration back to pH:

pH = -log([H+] average)

Simplifying the equation, we get:

pH = -log((10^-10 + 10^-12) / 2)

Calculating this expression, we find that the pH of the resulting solution is approximately 11.

Therefore, when equal volumes of solutions with pH 10 and pH 12 are mixed, the resulting solution will have a pH of 11.