First calculate the moles of Hcl= 0.154*10/1000=1.54x10-3 then calculate the new concentration of Hcl after adding water= 1.54x10-3/(990+10)*1000=1.54x10-3 then caluclate pH= -log(1.54x10-3)=2.81
First calculate the moles of Hcl= 0.154*10/1000=1.54x10-3 then calculate the new concentration of Hcl after adding water= 1.54x10-3/(990+10)*1000=1.54x10-3 then caluclate pH= -log(1.54x10-3)=2.81
dividing with 0.1 deffo wont give u the same number?
Exactly, as the answers has been posted above, just remember that the general way to do these questions is to calculate the amount of moles of the acid in the beginning, then calculate the concentration of the acid for the total volume, then find the pH as you would normally
You know the concentration of H+ in water at 298k is around 10−7M, using n=cv you can work out the moles of H+ in the water. Then you want to work out the moles of H+ added to the solution. Then use pH=−log[H+].
Well there isn't really a concentration of water, so adding some acid to water would still just make a dilute acid
Yeah I know, this is why what you said doesn't make much sense, the acid gets more dilute, but you can't say it gets weaker. pH does not really tell you how strong an acid is unless you're comparing 2 acids with same conc.
Yeah I know, this is why what you said doesn't make much sense, the acid gets more dilute, but you can't say it gets weaker. pH does not really tell you how strong an acid is unless you're comparing 2 acids with same conc.
Ah okay so it was just my wording was a bit off, thanks for letting me know so I can stop sounding like a fool