Zangetsu Posted April 24, 2016 Report Share Posted April 24, 2016 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? Reply Link to post Share on other sites More sharing options...
Emmi Posted April 24, 2016 Report Share Posted April 24, 2016 (edited) 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 April 24, 2016 by Emmi Reply Link to post Share on other sites More sharing options...
Msj Chem Posted April 25, 2016 Report Share Posted April 25, 2016 (edited) 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 April 25, 2016 by Msj Chem Reply Link to post Share on other sites More sharing options...
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