What is the pH of a 0.1 M HCl solution?
pH = 1
HCl is a strong acid so fully dissociates, so the concentration of H+ is equal to that of the HCl and pH=-log[H+]
4.01 is the amount of pH in a 1.5 M HCl solution.
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
- log(0.00450 M HCl) = 2.3 pH =======
- log(0.34 M HCl) = 0.5 pH =======
You mean 0.1 M HCl? -log(0.1 M HCl) = 1 pH
- log(0.100 M HCl) = 1 pH =====
Its pH value is 2.
pH = 0.0 (zero)
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… Read More
- log(1 X 10 -4 M HCl) = 4 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
It doesn't matter how much you have, HCl (Hydrochloric Acid) has a pH of 1
Since HCl is a strong acid .00001M HCl will produce a .00001M hydrogen ion concentration. Therefore, the pH will be 5.
pH= -log [H+] = -log  = 0
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 of 0.0000001M hcl soln = 5.80
What is the pH of a solution prepared by diluting 3.0 ml of 2.5 M HCl to a final volume of 100mL with water?
.003 L HCL * 2.5 moles HCL/1L = .0075 moles HCL .0075 moles HCL/ .1 L HCL = .075 M HCL = .075 M H3O+ pH = -log[H3O+] pH = -log[.075] = 1.12 this of course is not adjusted for activities of the ions in solution, but it would be very close to this value.
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
4 - LOG10 10-4
HCl 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)
Find moles HCl. 5 g HCl (1 mole HCl/36.450 grams) = 0.1372 moles HCl Now, Molarity = moles of solute/Liters of solution Molarity = 0.1372 moles HCl/1 liter = 0.1372 M HCl Then. -log(0.1372 M HCl) = 0.9 pH ( you might call it 1, but pH can be off the scale ) -----------
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
- log(0.0235 M HCl) = 1.6 pH =============If I remember correctly two places are a pH designation standard.
0.25 M HCl - log(0.25 M HCl) = 0.6 pH
0.125 moles HCl/1 liter = 0.125 M HCl HCl dissociates completely, so [H+] = 0.125 M pH = -log [H+] = -log 0.125 pH = 0.903
pH of 3.00 M of HCl
dilute your HCl solution to 0.2 M HCl solution and then follow above mentioned link : http://delloyd.50megs.com/moreinfo/buffers2.html
25g HCl 1 mol 36.46g HCl =.686 mol M=.686 mol/1.5 L=.457M pH= -log(.457) pH= .34
I assume you mean, 0.004 M HCl -log(0.004 M HCl) = 2.4 pH ----------------
The concentration of H+ [H+] = 0.01 m/l by definition pH = -(log10 (0.01)) therefore pH = 2
pH=-lg[H+] [H+]=10-pH With pH=2.0: Corrected: [H+]= 10-pH = 10-2.0 = 0.010 M HCL
p(x)=-log([x]) So... pH=-log([H+]) pH = 1.4
.260 M of HCL, not 260 More than likely correct, but, - log(0.260 M HCl) = 0.6 pH ----------- ( pH can be below 1 )
- log(0.25 M HCl) = 0.6 pH ------------
4nM HCl = 0.004 M -log(0.004 M) = 2.4 pH
For pH = 7,0: 756 mL disodium hydrogen phosphate (Na2HPO4) solution 0,1 M (14,2 g/L) + 244 mL hydrochloric acid (HCl) solution 0,1 M
How to prepare 2M HCl solution from con HCl
How does the pH level of a dilute solution of HCI compare with the pH level of concentrated solution of the same acid?
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.
By definition [H+]=10-pH so pH = -log[H+]. So for 10 M HCl >>> pH = -log(10) = -1.0. This is only of theoretical importance because pH formula are for ideal solutions, which are only valid for diluted, not concentrated solutions. Also a pH-meter will not give relyable measurements at pH << 1.0 or below.
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.
0.1 M HCl =============
Acids have pH value less than 7 and bases higher , so NaOH has higher pH value.
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]
What is the molarity of an hcl solution if 7ml hcl solution is titrated with 27.6ml of 0.170m of naoh solution?
(7 mL)(X M HCl) = (27.6 mL)(0.170 M NaOH) 7X = 4.692 X = 0.7 M HCl ==========
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… Read More
About pH= 2.4 for 1.0 M solution, pH= 2.9 for 0.10 M solution, pH= 3.4 for 0.010 M solution
Is 2.77 a normal pH for hydrochloric acid thats 1M If not then how should I calibrate my pH meter at home?
A pH meter is calibrated with a standard solution of pH, having adequate electrodes. Read also the manual of the instrument; apply the correction for temperature if necessary, etc.
Calculate the pH of a solution prepared by diluting 3.0 mL of 2.5 M HCl to a final volume of 100 mL with H2O?
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)
pH = -log[H+] That is, the pH of a solution is the negative log of the concentration of hydrogen atoms. The concentration of hydrogen atoms must be in units of Molarity, or moles per liter. In order to determine the pH of a solution containing 1 mole of HCl, you must also know the volume of the solution. I am assuming the question is: What is the pH of 1M HCl, and not 1 mole… Read More
0.25 N HCl is the same as 0.25 M HCl which is 0.25 moles HCl/liter of solution.