What is the pH of 10 M of H+?
921604
theoretically yes 10m acid has a pH of -1, and 10 moles of base pH 15.\n\nBut , in fact, for very high concentrations, you must not take in account the concentration of ions H+ but the activity . The actvity takes in account the high concentration and is \n\nactivity = gamma * pH where gamma is lower than one for high concentrations
249301
H3O+ .0292M = pH 1.53
860365
use the equation  pH= -lgKa+ lg( salt/acid)\n(this equation only applies for buffer system)
696206
the moles of NaOH = the moles of OH - = 0.125 * 0.606 = 0.07575\nthe moles of OH - in the new solution = 0.07575/(15 + 0.125 ) = 5*10^(-3) M\npOH = - log ( 5*10^ -3) = 2.3\n=> pH = 14 - pOH = 11.7
907196
The concentration of H+ is 3.0 x 10^-4, so you know that the pH will be around 4.  What you need is a calculator that has a LOG button to convert the 0.0003 to an integer value.\n\nSo on your calculator you press the LOG button and then type in 0.0003 and then a ")" and press ENTER.  \n\nYou should get a result that says: -3.522878745 or so.  You need to round off to -3.5 and since pH is the NEGATIVE log of the hydrogen ion concentration you get rid of the minus.\n\nThe answer is 3.5.\n\n1x10^-3.5 is approximately equal to 0.0003.
272078
I take it you mean the concentration of hydronium ions (H3O+) at a pH of 6.2. ok, the equation you need is this:\npH = -log [H+]\n\nTo solve, do this:\npH = -log [H+]\n-pH = log [H+]\n10 ^ (-pH) = [H+]   (if you don't understand this step, look up logarithms. Basically, a= log b means exactly the same thing as 10^(a)=b )\n\n10^(-6.2) = [H+]\nYou have 10^(-6.2) mol/ L of hydronium ions.\n\nAnother useful equation is this:\n[H30+] * [OH-] = 10^-14\nwhich will give you the concentration of OH- ions when you have the concentration of hydronium ions, or vice versa.
350659
Why do people answer complete and total bullcrap answers?\n\n\nThe only thing they were right about was it is a weak acid, and even that bears no relavance to its P.H only you use a diferent formula in calculating it\n\n\nthe question is incomplete, you need to know the molarity of the acid and then you get the square root of it disociation constant (Ka) multiplied by its molarity\n\nand then you -log(base10)the answer you got from the first part and that is the P.H\n\nin short  Sq root(Ka*molarity acid)= H+ concentration\n\nP.H= -logH+
148348
Perhaps a little less than pH=7. You can try pH=6, due to acidic rain (decreases the pH of lake), then again it really depends on the size of the lake, because of the buffer solution (ionic equilibria in chemistry) present in the lake and all.\n\nIf you know the concentration of H+ ions in the lake, all the better. The pH will be: pH=-log[H+].
553835
Well to determine [H+] you use the equation that \npH=-log [H+] \nthus a solution with pH=3 will have a [H+] of 1x10^-3, \nand a solution with pH=8 will have a [H+] of 1x10^-8\nTo find [OH-] there are several equations you can use\n[OH-]=Kw/[H+], \nKw is the ionization constant of water which is equal to 1.0x10^-14\nThus you will have [OH-] of 1.0x10^-11 and 1.0x10^-6\nSo you answer will be that the solution with pH=3 will have a [H+]>[OH-]
768732
pH at pKa +- 1 unit. from henderson hasselbalch\npH = pKa + log [base]/[acid] \ngiving +- 1 when the ratios are 1/10 or 10/1 in concentration\nthe capacity of a buffer is achieved when the ratio of weak acid to conjugate base exceeds 10:1 or vice versa. \nDoes that help?
685262
the answer depends on how many H+ need neutralising by the NaOH.\nassuming it is only 1 per molecule of the acid then you can use the following equations:\n\nno.of moles = (vol/1000)*concn\nno. of moles = (40.11/1000)*0.240 = 0.0096264\n\nno. of moles = mass(g) / RMM\ntherefore\n0.0096264 = 0.55 / RMM\ntherefore RMM = 0.55 / 0.0096264\nRMM = 57.134\n\nif you have 2H+ that need neutralizing then the molecular weight = 2*57.134 = 114.268 etc etc
224213
well, pH = -log[H+]. so if you just undo the log you will have the concentration of H+ (which is the same as H30+). then .. [H+} + [OH-] = 14 good luck!
1066669
we know that pH is ranged from totally basic to totally acidic, with neutrality (water) being at pH=7; so just display a range from 0 to 14 and color code it (red for acid and blue for basic) and place your findings there (ie before and after situations of NaCl towards the pH of the solution). Also place analogies (ie. the results of the pH due to NaCl concentration of 10% is equivalent to household ammonia)\n\nWhen explaining something to a non-science person, visuals and analogies always helps.
1167469
pH is a measure of the activity of hydrogen ions (H+) in a solution and, therefore, its acidity or alkalinity. For dilute solutions, however, it is convenient to substitute the activity of the hydrogen ions with the molarity (mol/L) of the hydrogen ions (however, this is not necessarily accurate at higher concentrations [1] [2]).\n\nIn aqueous systems, the hydrogen ion activity is dictated by the dissociation constant of water (Kw = 1.011 × 10−14 M2 at 25 °C) and interactions with other ions in solution. Due to this dissociation constant a neutral solution (hydrogen ion activity equals hydroxide ion activity) has a pH of approximately 7. Aqueous solutions with pH values lower than 7 are considered acidic, while pH values higher than 7 are considered alkaline.\n\nThe concept was introduced by S.P.L. Sørensen in 1909, and is purported to mean "pondus hydrogenii" in Latin.[3] However, most other sources attribute the name to the French term pouvoir hydrogène. [4] [5] [6] In English, pH can stand for "hydrogen power,"[4] [5] [6] "power of hydrogen," [7] [8] or "potential of hydrogen."[3] [9] [10] All of these terms are technically correct.\nThough a pH value has no unit, it is not an arbitrary scale; the number arises from a definition based on the activity of hydrogen ions in the solution.\n\nThe precise formula for calculating pH is:\n\n \naH+ denotes the activity of H+ ions, and is unitless.\n\nIn dilute solutions (such as tap water), the activity is approximately equal to the numeric value of the concentration of the H+ ion, denoted as [H+] (or more accurately written, [H3O+]), measured in moles per litre (also known as molarity). Therefore, it is often convenient to define pH as:\n\n \nFor both definitions, log10 denotes the base-10 logarithm, therefore pH defines a logarithmic scale of acidity. The straight bars, indicating absolute value, make pH a unit-less quantity. For example, if one makes a lemonade with a H+ concentration of 0.0050 moles per litre, its pH would be:\n\n \n \nA solution of pH = 8.2 will have an [H+] concentration of 10−8.2 mol/L, or about 6.31 × 10−9 mol/L. Thus, its hydrogen activity aH+ is around 6.31 × 10-9. A solution with an [H+] concentration of 4.5 × 10−4 mol/L will have a pH value of 3.35.\n\nIn solution at 25 °C, a pH of 7 indicates neutrality (i.e. the pH of pure water) because water naturally dissociates into H+ and OH− ions with equal concentrations of 1×10−7 mol/L. A lower pH value (for example pH 3) indicates increasing strength of acidity, and a higher pH value (for example pH 11) indicates increasing strength of alkalinity. Note, however, that pure water, when exposed to the atmosphere, will take in carbon dioxide, some of which reacts with water to form carbonic acid and H+, thereby lowering the pH to about 5.7.\n\nNeutral pH 25 °C at is not exactly 7. pH is an experimental value, so it has an associated error. Since the dissociation constant of water is (1.011±0.005)×1014, pH of water at 25 °C would be 6.998±0.001. The value is consistent, however, with neutral pH being 7.00 to two significant figures, which is near enough for most people to assume that it is exactly 7. The pH of water gets smaller with higher temperatures. For example, at 50 °C, pH of water is 6.55±0.01. This means that a diluted solution is neutral at 50 °C when its pH is around 6.55 and that a pH of 7.00 is basic.\n\nMost substances have a pH in the range 0 to 14, although extremely acidic or extremely basic substances may have pH less than 0 or greater than 14. An example is acid mine runoff, with a molar concentration of 3981 M and a pH = -3.6.\nThe Grand Peptide Families\nThese peptides are ribosomal peptides, usually with hormonal activity. All of these peptides are synthesized by cells as longer "propeptides" or "proproteins" and truncated prior to exiting the cell. They are released into the bloodstream where they perform
834164
Mg(OH)2  --->  Mg^2+  +  2OH^-\n\nk(sp) = [Mg^2+][OH^-]^2 \n\nIf pH = 12.8, then\npOH = 14 -12.8 = 1.2\ntherefore, [OH^-] = 10^1.2 =  0.0631 (ask your calculator!)\nnow, k(sp) = 1.2 * 10^-11 = [Mg^2+] [0.0631]^2 \nIf you solve for [Mg^2+] , you will have the answer.\n\nIf you need more help, please email me or re post.
1125360
H2O2 can exist in many forms at pH = 7, \n\nH3O2+, H3O3- are among the possibilities.
957414
P4 O 10 + 6 * H2O = 4 H3 PO4