# Acids and bases, is the correct way to solve this problem?

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In a practice question I was doing, it says that 10 cm^3 of 0.01 mol dm^3 of HCL is diluted with 90 cm^3 water and the questions ask to calculate the ph. So if I am correct, the starting PH is 2, due to 0.01 mol being equal to 10^-2. 2nd, with the 10 cm^3 of acid being diluted with 90 cm^3 of water, then would that cause the resulting ph to be a value of 3 as the water dilution makes the acid less acidic?

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Yes. When you convert your volume to dm^3 and use the formula for dilution, you'll find that your final concentration is 0.001 mol/dm^3. Since HCl is a strong acid, we find the pH by -log(0.001), which is 3.

Edited by Emmi

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Basically you are diluting the acid by a factor of 10. pH is an inverse and logarithmic scale so a decrease in the [H+] by a factor of 10 results in an increase of one pH unit. Therefore the pH changes from 2 to 3.

Maybe this video will help:

Edited by Msj Chem

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