HNO3 is a strong acid, so this is fairly straight forward question.
Calculate the new molarity of the acid after being mixed with H2O:
0.01M*0.005L= moles of HNO3.
moles of HNO3/(0.005L+0.05L)= Molarity of HNO3 after addition of water.
Since this is a strong acid, you can just use the following equation:
pH=-log[H^+]
where
[H^+]=Molarity of HNO3 after addition of water.
****Note: Volumes are not additive, but you should get the correct result-- I believe.
calculate the change in pH when 5 ml of 0.01M HNO3 is added to 50 ml of water.
I know how to calculate pH, but not when it is changing.
3 answers
Devron is right; volumes are not additive but at these concentrations the difference is strictly theoretical. I don't think you would ever be able to measure it.
Thank you, I was totally over thinking the problem
3 answers