A-level chemistry help urgenttt!!!Watch this thread
Calculate the PH of the solutions formed in the following way.
Addition of 250 cm3 of water to 50cm3 of 0.2 moldm-3 HN03
Someone please help😂
Sorry I'm just trying to understand
calculate the change in the concentration of HNO3. now that you have a new concentration and HNO3 is a strong acid, the pH of the solution is the negative log (to the base ten) of the concentration of H+ ions, which, in this case, is the same as the concentration of the acid
To calculate the ph of 10 gdm-3 Hcl
So I have done: 10/36.5 = 0.27
-log(0.27) = 0.56
Is this correct?
I had the same question and that is what I did!