Approximately how much water should be added to 10.0 mL of 10.3 M HCl so that it has the same pH 0.90 M acetic?

A) Approximately how much water should be added to 10.0 mL of 10.3 M HCl so that it has the same pH 0.90 M acetic acid (Ka = 1.8 x 10^-5)?



B) What volume of water must be added to 14.6 mL of a pH 2.0 solution of HNO3 in order to change the pH to 4.0?



If you could explain in detail so I can remember th steps that are needed to solve other familiar problems. Thank you for your help and making it easier for me to learn from Chemistry.Approximately how much water should be added to 10.0 mL of 10.3 M HCl so that it has the same pH 0.90 M acetic?
Lets not use pH, and leave it as the simpler [H+]



[H+] of 0.9 M HA (Ka = 1.8x10^-5) is what?





That would be 0.0040159324 M [H+]





Now, (0.01 * 10.3 M) / (0.01+x) = 0.004159324 (0.12 moles H+ divided by

the volume of water - total volume, because you already have 0.01

Liters)

x = 29.871 Liters