pH = 5.56 at 36.0 ml(equivalence point)
for 1L 121.1 gr tris base 80 ml HcL 920 ml Distilled Water
The answer depends on several unspecified variables, most importantly the final molarity of the solution, which depends on the final volume. You can calculate the value yourself using the formula: pH = -log[H+] where [H+] is the final concentration of H+ ions in solution. For HCl, [H+] is equal to molarity. So, for example, if you add 50.0 ml of 1.0M HCl to 950 ml of deionized water, your final concentation is: (50.0 ml/1000 ml) * (1.0M) = 0.05M Therefore: pH = -log[0.05] = 1.3
For example, to obtain a solution with the pH=7,00 mix: 756 mL 0,1 M solution of Na2HPO4 with 244 mL of 0,1 M HCl solution.
pH is an INTENSIVE property as it does not depend on the amount of material present. The pH of 1 ml of solution x is the same as the pH of 100 ml of solution x.
Because HCl is a strong acid, it dissociates completely to H + Cl. Therefore, 3.0 mL x 2.5 M HCl = 7.5 meq of H+ . In 100 mL of solution, this is 0.075 M H+. pH= - log [H] = - log (0.075) = - (-1.1) = 1.1 (two significant figures)
2.7
pH = 5.56 at 36.0 ml(equivalence point)
for 1L 121.1 gr tris base 80 ml HcL 920 ml Distilled Water
Volumetric flask= 200 ml (100 ml K2HPO4 0.1M)+ (44.6 ml HCl 0.1M) added in flask then added deionized or distilled water until mark.
The answer depends on several unspecified variables, most importantly the final molarity of the solution, which depends on the final volume. You can calculate the value yourself using the formula: pH = -log[H+] where [H+] is the final concentration of H+ ions in solution. For HCl, [H+] is equal to molarity. So, for example, if you add 50.0 ml of 1.0M HCl to 950 ml of deionized water, your final concentation is: (50.0 ml/1000 ml) * (1.0M) = 0.05M Therefore: pH = -log[0.05] = 1.3
For example, to obtain a solution with the pH=7,00 mix: 756 mL 0,1 M solution of Na2HPO4 with 244 mL of 0,1 M HCl solution.
pH is an INTENSIVE property as it does not depend on the amount of material present. The pH of 1 ml of solution x is the same as the pH of 100 ml of solution x.
- log(0.25 M HCl) = 0.6 pH ------------
- log(0.00450 M HCl)= 2.3 pH=======
I think it'd be pH 7. Same amount of both, providing they are the same molarity!
pH is 9.75