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?
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??
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?
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.
Yes, you are using the right steps to calculate the pH of HCl. The pH of a solution is found by taking the negative logarithm (base 10) of the concentration of hydrogen ions ([H+]) in the solution.
In the case of HCl with a concentration of 0.01 M, you correctly started by taking the negative logarithm of 0.01: pH = -log(0.01). When you enter this into a calculator, you should indeed get a pH value of 2.
So, you are using the correct formula and steps to calculate the pH of HCl.