HCl is a strong acid, therefore we can assume that it ionises completely into H+ (or H3O+ if you prefer) and Cl- ions.
Therefore an 0.10 M HCl solution will contain 0.10 moles/L of H+ ions.
pH = -log10[H+] where [ ] denotes "concentration of" in moles/L
Note that the formal definition of pH relies on the "activity" of H+ (which is the product of an activity coefficient and concentration), but for simple problems this can be be ignored.
So pH = -log[0.1]
= 1.0
The pH of a 0.280 M HCl solution is approximately 0.55. This is because HCl is a strong acid that dissociates completely in solution to produce H+ ions, leading to a low pH value.
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
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)
p(x)=-log([x])So... pH=-log([H+])pH = 1.4
Adding 6 M HCl (hydrochloric acid) lowers the solution's pH, causing a shift in methyl red's equilibrium towards the red form. This is because methyl red is a pH indicator that changes color from yellow (at a higher pH) to red (at a lower pH). The addition of acid protonates the indicator, resulting in the color change.
The pH of a 6M HCl solution is 0.
The pH of a 1N HCl solution can be calculated using the formula pH = -log[H+], where [H+] is the concentration of H+ ions in the solution. For a 1N solution of HCl, the concentration of H+ ions will be 1M, so the pH will be -log(1) = 0.
The pH of a solution containing 6M HCl is 0.
To calculate the pH of the resulting solution when mixing HCl and NaOH, you need to determine the moles of each reactant. Use these values to find the limiting reagent and the excess reagent. Then, calculate the concentration of the excess reagent left in solution. Finally, use the concentrations of the excess reagent and water to determine the pH of the resulting solution.
The pH of a solution of HCl is calculated using the concentration of H+ ions. Since HCl dissociates completely in water, the concentration of H+ ions is equal to the concentration of HCl. First, calculate the concentration of HCl in the solution: 0.45g HCl / (36.46 g/mol) = 0.012 mol HCl. Then, divide the moles of HCl by the volume of the solution in liters to get the concentration in mol/L. Finally, calculate the pH using the formula pH = -log[H+].
To calculate the pH of a 0.001 mol/dm³ HCl solution, you use the formula pH = -log[H+]. For HCl, since it is a strong acid, it dissociates completely into H⁺ ions. Thus, the concentration of H⁺ ions in a 0.001 mol/dm³ HCl solution is also 0.001 mol/dm³. Taking the negative logarithm of 0.001 gives a pH of 3.
- log(0.00450 M HCl)= 2.3 pH=======
HCl is a strong acid and dissociates completely. Therefore it can be found using the equation: ph= -log [H+]
The pH of a 0.0001M aqueous solution of HCl is 4. The pH of a solution is calculated using the formula pH = -log[H+], where [H+] is the concentration of hydrogen ions in the solution. Since HCl is a strong acid that dissociates completely in water, the concentration of H+ ions in a 0.0001M solution of HCl is also 0.0001M.
The pH of a 42m HCl solution would be approximately -log(42) = -1.62. This solution is strongly acidic.
To calculate the molarity of a solution from its pH, use the formula: pH = -log[H+]. In this case, pH 5.7 corresponds to [H+] = 1 x 10^-5.7 M. Given that HCl is a strong acid and dissociates completely in water, the molarity of HCl is also 1 x 10^-5.7 M.
The pH of a 0.280 M HCl solution is approximately 0.55. This is because HCl is a strong acid that dissociates completely in solution to produce H+ ions, leading to a low pH value.