pH = -log10[H+], where [H+] is the hydrogen ion concentration.
So, in this case, pH = -log10[1], yielding pH = 0.
1N - conc will change the pH too quickly. You might want to consider 0.1molar for finer adjustments.
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
Its pH value is 2.
- log(0.00450 M HCl)= 2.3 pH=======
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
1N - conc will change the pH too quickly. You might want to consider 0.1molar for finer adjustments.
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
HCl is a strong acid and dissociates completely. Therefore it can be found using the equation: ph= -log [H+]
its PH is 3
Its pH value is 2.
- log(0.00450 M HCl)= 2.3 pH=======
if its complete dissociation, then the products would be a salt and water, which means the pH is 7 or neutral. OMG, if the pH is currently 4 then [H+] = 1.0 e-4 M (pH = -log[H+]) if you add 0.003 moles then 1.0e-4 M +.003 M = .0031 M (Since the strong acid HCL completely dissociates in aq solution) pH = -log [.0031M] = 2.51
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.
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.
HCl is a strong acid, so we assume that it completely breaks up into ions in solution. HCl ----> H+ & Cl- if we have 0.01m of HCl, it will give 0.01m of H+ and 0.01m Cl- pH = -log [H+] pH = -log 0.01 pH = 2
pH = -log(0.280) = 0.553