0
HCl liberates 1M of H+ Ions per mole of HCl so
0.034M HCl = 0.034 M H+ Ions
as pH = -log10 [H+]
where [] means the conc.
pH= -log10 [0.034]
Wiki User
∙ 9y ago
Wiki User
∙ 14y ago3
We know that "pH = -log (H+)"
If the concentration of HCl is 0,001M the concentration of H+ is also 0,001M. So "-log 0,001 = 3". The answer is 3.
Wiki User
∙ 11y ago- log(0.34 M HCl)
= 0.5 pH
=======
Wiki User
∙ 6y agoHCl is a strong acid so 0.035 M HCl ==> 0.035 M H^+.pH = -log [H+] = -log 0.035
pH = 1.46 = 1.5 (2 sig. figs)
Wiki User
∙ 11y ago2.4
Wiki User
∙ 14y ago1.5
Diego Borges
09
Anonymous
1.5 is the right answer
Anonymous
0.30
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The pH of a 0.01 M solution of HCl in water would be?
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
What is the pH of a 1.0x10-4m hcl solution?
- log(0.00450 M HCl)= 2.3 pH=======
What is the pH of a solution of 0.010 M HCl?
Its pH value is 2.
What is theoretical pH value for 1 m hcl solution?
It doesn't matter how much you have, HCl (Hydrochloric Acid) has a pH of 1
What is the pH of 3.0 M KOH?
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
The pH of a 0.01 M solution of HCl in water would be?
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
What is the pH of a 0.001 M of HCl solution?
its PH is 3
What is the pH of a 1.0x10-4m hcl solution?
- log(0.00450 M HCl)= 2.3 pH=======
What is the pH of a solution of 0.010 M HCl?
Its pH value is 2.
What is theoretical pH value for 1 m hcl solution?
It doesn't matter how much you have, HCl (Hydrochloric Acid) has a pH of 1
What is the pH of 3.0 M KOH?
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
What is the pH of 10.0L of 0.0020 M HCL?
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
What is the pH of Na2HPO4?
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.
What is the pH of a 1.8 x 10-5 M HCl Solution?
25g HCl 1 mol 36.46g HCl =.686 mol M=.686 mol/1.5 L=.457M pH= -log(.457) pH= .34
What is the pH of a solution that contains 0.10 mole of HCI in a voulme of 100.0l?
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
What is the pH of a 0.0235 M HCI solution?
- log(0.0235 M HCl)= 1.6 pH=============If I remember correctly two places are a pH designation standard.
The pH in aqueous solution of hcl at 25c is 4 hcl0.003 moles was added to 1l of this solution calculate the new pH in the solution assume complete dissociation of hcl?
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
