Calculate the pH of:0.01M HCl pH=-log[H*]. ..=-log[0.01]. ..=2.is the pH of HCl,I enter -log on my calculator Igot 2 as my answer,Am I using the right step of formular?

Answer 1

0.01M HCl.................... pH=-log[H*]........
..=-log[0.01]................
..=2. I used my calculator to enter the above problem and my answer is 2.Am I using the right formular??I'm not sure about that formular because of pH14 infront of a solution,why is pH 14 infront of the poH??

Answer 2

pH=-log[H*].
=-log[0.01]
=2

I've used my calculator by entering -log in brackets 0.01 and I got 2 as my answer,Am I using the right step of formular?I'm not sure about my answer because of pH 14 in front of a solution of hydrogen,ph=-log[H*]
poH=14-pH
Could someone please show me why they have to put that 14 infront? Is it for subtracting the above question?

Answer 3

I'm not sure what's going on. Yes, pH of 0.01M HCl is 2 and you're doing that right on the calculator to get that answer. As for the other. if you want the pOH of a solution, which seems to be another separate problem, you use
pH + pOH = pKw
pKw is 14 so pOH = 14-pH
pOH = -log(OH^-) also just as pH = -log(H^+) and pKw = -logKw and pKa = -logKa or pKb = -logKb.