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25g HCl 1 mol 36.46g HCl =.686 mol M=.686 mol/1.5 L=.457M pH= -log(.457) pH= .34
No, pH 2.77 is not the correct pH for 1 M HCl. The pH of 1 M HCl should be 0 (zero) because pH is the negative log the the H+ and for 1 M HCl the [H+] is 1 M, and the negative log of 1 is 0.
A solution of HCl is highly dissociated into ions, A 0.000001 M solution (1 x 10-6) has a pH of 6 ... close to neutral. A 0.001 M solution (1 x 10-3) has a pH of 3 ... more concenterated, but still not a really concentrated solution. A 0.1 M solution (1 x 10-1) has a pH of 1 ... even more concentrated. showing it is more acidic.
Yes, 0.1 M HCl has pH value of 1.0, 1.0 M HCl has pH value of 0.0, 10.0 M pH = -1.0 (even negative is possible, but this is about the limit)
- log(0.2 M) = 0.70 pH =======
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
its PH is 3
- log(0.00450 M HCl)= 2.3 pH=======
Its pH value is 2.
It doesn't matter how much you have, HCl (Hydrochloric Acid) has a pH of 1
Molarity is the concentration of a solution, defined as moles per unit volume. Where, Molarity = moles / volume In this case the molarity of the HCl solution is 0.03 M The pH of this is calculated by the equation below pH = - log [H+] Where [H+] is the concentration of hydrogen ions/ protons present in the solution. As HCl only contains one hydrogen ion per molecule then the concentation of [H+] is 0.03 M Then the equation can be calculated (the minus sign is very important!) pH = - log 0.03 pH = 1.52 To summarise, Molarity of 0.03 M HCl solution is 0.03 M pH of 0.03 M HCl solution is 1.52
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
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.
25g HCl 1 mol 36.46g HCl =.686 mol M=.686 mol/1.5 L=.457M pH= -log(.457) pH= .34
100 Liters? I will assume as much. Molarity = moles of solute/Liters of solution Molarity = 0.10 mole HCl/100.0 Liters = 0.001 M HCl -------------------------now, to find pH - log(0.001 M HCl) = 3 pH -----------------so, your acid is of 3 pH, which is to be expected at the volume od solution
- log(0.0235 M HCl)= 1.6 pH=============If I remember correctly two places are a pH designation standard.
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