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
- log(0.00450 M HCl)= 2.3 pH=======
Since HCl is a strong acid it completely dissociates. Therefore [H+] = [HCl] and this case = 0.25 M. pH = -log [H+] = 0.602
- log(0.25 M HCl) = 0.6 pH ------------
.260 M of HCL, not 260 More than likely correct, but, - log(0.260 M HCl) = 0.6 pH ----------- ( pH can be below 1 )
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
- log(0.00450 M HCl)= 2.3 pH=======
Since HCl is a strong acid it completely dissociates. Therefore [H+] = [HCl] and this case = 0.25 M. pH = -log [H+] = 0.602
- log(0.25 M HCl) = 0.6 pH ------------
.260 M of HCL, not 260 More than likely correct, but, - log(0.260 M HCl) = 0.6 pH ----------- ( pH can be below 1 )
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
pH=-lg[H+][H+]=10-pHWith pH=2.0:Corrected:[H+]= 10-pH = 10-2.0 = 0.010 M HCL
0.1 M HCl =============
In 0.01 M of HCl, the concentration of the Hydronium ions is 0.01M as well since HCl is monoprotic. pH = -log [H3O+] = -log 0.01 = -log10-2 = -(-2log10) = 2 Thus, the pH of 0.01 M HCl is 2.
its PH is 3
Its pH value is 2.
HCl liberates 1M of H+ Ions per mole of HCl so 0.034M HCl = 0.034 M H+ Ions as pH = -log10 [H+] where [] means the conc. pH= -log10 [0.034]
25g HCl 1 mol 36.46g HCl =.686 mol M=.686 mol/1.5 L=.457M pH= -log(.457) pH= .34