pH=-lg[H+]
[H+]=10-pH
With pH=2.0:
Corrected:
[H+]= 10-pH = 10-2.0 = 0.010 M HCL
The pH of a 6M HCl solution is 0.
- log(0.00450 M HCl)= 2.3 pH=======
The pH of a solution containing 6M HCl is 0.
The pH of water decreases after adding HCl due to the increase in hydrogen ions. The resulting pH level depends on the amount of HCl added.
.260 M of HCL, not 260 More than likely correct, but, - log(0.260 M HCl) = 0.6 pH ----------- ( pH can be below 1 )
The pH of a 6M HCl solution is 0.
- log(0.00450 M HCl)= 2.3 pH=======
The pH of a solution containing 6M HCl is 0.
The pH of water decreases after adding HCl due to the increase in hydrogen ions. The resulting pH level depends on the amount of HCl added.
A 0.1 M concentration of HCl corresponds to a pH of 1.0.
.260 M of HCL, not 260 More than likely correct, but, - log(0.260 M HCl) = 0.6 pH ----------- ( pH can be below 1 )
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.
I think it'd be pH 7. Same amount of both, providing they are the same molarity!
The pH of a 0.25 M HCl solution is approximately 0.60. This is because HCl is a strong acid that dissociates completely in water to form H+ ions, resulting in a high concentration of H+ ions and a low pH.
The pH of a 0.005N HCl solution is approximately 2.3. This is because HCl is a strong acid that dissociates completely in water to form H+ ions, leading to an acidic pH.
Since HCl is a strong acid it completely dissociates. Therefore [H+] = [HCl] and this case = 0.25 M. pH = -log [H+] = 0.602
its PH is 3