- log(0.00450 M HCl)
= 2.3 pH
=======
I assume you mean,
0.004 M HCl
-log(0.004 M HCl)
= 2.4 pH
----------------
The pH is four (4).
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
Its pH value is 2.
0.002M HCl means 0.002 moles HCl in 1L solution. Therefore 0.02 moles HCl in 10L solution. pH = 2-log2 = 2-0.3010 = 1.6990
In solution with a pH of 1 [H+] is 0.1M. Since HCl is a strong acid [HCl] will also be 0.1M. So, in 1 liter of solution you will have 0.1 mol of HCl.
the pH is .377 the pH is .377
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
its PH is 3
Its pH value is 2.
0.002M HCl means 0.002 moles HCl in 1L solution. Therefore 0.02 moles HCl in 10L solution. pH = 2-log2 = 2-0.3010 = 1.6990
In solution with a pH of 1 [H+] is 0.1M. Since HCl is a strong acid [HCl] will also be 0.1M. So, in 1 liter of solution you will have 0.1 mol of HCl.
pH = -log(0.280) = 0.553
the pH is .377 the pH is .377
It doesn't matter how much you have, HCl (Hydrochloric Acid) has a pH of 1
HCl
100 Liters? I will assume as much. Molarity = moles of solute/Liters of solution Molarity = 0.10 mole HCl/100.0 Liters = 0.001 M HCl -------------------------now, to find pH - log(0.001 M HCl) = 3 pH -----------------so, your acid is of 3 pH, which is to be expected at the volume od solution
7
HCl is a strong acid. Therefore, it can be expected to fully dissociate in aqueous solution, yielding one hydrogen ion and one chloride ion per molecule. The concentration of the hydrogen ion should thus be the same as the initial concentration of the HCl. Therefore, a 0.10M HCl solution has an H+ concentration of 0.10M. By the equation pH=-log[H+], the pH of this solution is 1.