- log(0.25 M HCl) = 0.6 pH ------------
- log(0.00450 M HCl)= 2.3 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=-lg[H+][H+]=10-pHWith pH=2.0:Corrected:[H+]= 10-pH = 10-2.0 = 0.010 M HCL
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 ------------
- log(0.00450 M HCl)= 2.3 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=-lg[H+][H+]=10-pHWith pH=2.0:Corrected:[H+]= 10-pH = 10-2.0 = 0.010 M HCL
Since HCl is a strong acid it completely dissociates. Therefore [H+] = [HCl] and this case = 0.25 M. pH = -log [H+] = 0.602
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.
0.1 M HCl =============
The pH increases because the HCl is becoming less acidic. A pH of 7 is neutral. A pH falls below 7, acidity increases. As pH rises above 7, basicity increases. Diluting HCl means that the HCl becomes less concentrated, and therefore, less acidic. As it becomes less acidic, the pH will become more basic, and thus increase.
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
It solely depends on H+ concentration: each HCl gives one H+ , to calculate use pH = -log[H+] So, at [HCl]=1.0 >> pH= 0.0 at [HCl]=0.5 >> pH= 0.7 at [HCl]=0.1 >> pH= 1.0 at [HCl]=1.0*10-5 >> pH= 5.0 but don't ever use this simplified 'acid pH' calculus method when the answer comes close to (or exceeds) 6.5, 7 or 8 etc.
its PH is 3
1.3