Different pH change magnitude question?
355477
Buffers resist change in pH. This is generally a result of the buffer containing a mix of a weak acid and the salt of the conjugate base. \n\nWhen you add NaOH to the buffer, it dissociates into Na+ and OH-. The OH- reacts with the H+ ions which the weak acid has dissociated into (along with the conjugate base) up to the acid dissociation constant, Ka, of the acid.\n\nThis shifts the equilibrium of the dissociation reaction of the acid, and more dissociates, maintaining a relatively stable pH, as compared to when the NaOH is added to the distilled water. (If you had instead added a base, it would instead have shifted the equilibrium of the base, and, instead, more of the base would have reacted with the free H+, again, keeping the pH relatively constant)
836698
The greater the Kb value is, the stronger the base.  Just think about what Kb means.  On the left side of the equilibrium is the base, on the right side is the conjugate acid, meaning the base has reacted.  The stronger the base, the more it will react, the more the reaction will proceed to the right, the greater Kb will be.\n\nSo.....\n\nC > E > B > A > D
482772
Ph is A measure of the acidity or alkalinity of a solution, numerically equal to 7 for neutral solutions, increasing with increasing alkalinity and decreasing with increasing acidity. The pH scale commonly in use ranges from 0 to 14.
245193
Use your buffering equation and plug in the numbers you know to get your answer.  You might have to do reverse calulations to find your answer.
1173010
If you add one molar **water** to the equation?  There would be no change, since the reaction is already occurring in water!\n\nWater, by the way, is by definition 55.56M -- there are 55.56 moles of water in every liter of water (1000 g/18 g/mole).
687922
Actually, if you mean strength as in acidity, a solution with a pH of 5 would be 10 times stronger than one with a pH of 6.\n\nHowever, both would be very weak acids, since neutral is about pH 7.
201798
-log[0.140] = pOH (rather than pH because NaOH is a base)\n\npH + pOH = 14\n\nOne you have calculated your pOH using the antilog of 0.140, subtract that answer from 14 to find your pH!
1181487
True the higher the pH the lower the concentrations
89429
That would depend on what the system is, what the current pH is, what you are trying to accomplish (create a certain buffer, etc.)\n\nIf the current pH is > 7.5 you would use an acid (HCl, HNO3, etc.) to lower the pH.  If the current pH is < 7.5 you would use a base (NaOH, NH4OH) to raise the pH.
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
866949
<--------- 7 ---------->   (pH)\nacid                base\n\nthe smaller pH the stronger acid\nthe bigger pH the stronger base\n\nso you should arrange them from the smallest pH to the biggest\nstomach content (1.6)\nsoda pop (3)\nmilk (6.5)\nfresh water (7, of course neutral)\nhuman blood (7.4)\nsea water (8.7)\ndrain cleaner (8.7)
650541
pH = -log ([H3O+])\npOH = -log ([OH-])\n\npH + pOH = pKw = 14\npOH = 14 - pH\n\n[OH-] = 10^-(pOH)\n\nPlug in and solve.
114804
No need to break it down to "molecular" other wise, try this: Sodium thiosulphate. Dont just use it coz they all vary in their Ph U know the basisity or acidity they concentrate.
184683
pH is the measure of acidity or basicity(alkalinity) of a solution. When pH increases, the solution is more alkaline and when pH decreases the solution is more acidic. A pH of 7 implies that the solution is neither basic(alkaline) nor acidic but neutral. Check the answer to one of the similar questions on Yahoo answers at the following page.
358982
NaOH is completely dissociated when it is dissolved in water.  A .0025M KOH solution therefore produces a [OH-] equal to .0025M.  The ion product of [H30+] and [OH-] is a constant, 1.0 x 10^-14 M^2.  By substitution, the [H30+] can be determined.  The pH can then be calculated.\n\n[H30+] = 1.0 x 10^-14 M^2/[OH-] = 1.0 x 10^-14 M^2/2.5 x 10^-3 \n\naaaand all that equals .4 x 10^-11... and you put that in correct scientific notation... 4 x 10^-12 M for the concentration of [H30+].\n\nWe then plug that in...   \n\npH= -log[H30+] = -log (4 x 10^-12) = 11.4\n\nThe pH of a .0025M solution of KOH is 11.4, which is greater than 7, making it a basic solution.\n\nStrong base at that.\n\nP.S.- If it is supposed to have the correct number of significant figures/digits, the answer would then be 11.
404745
The "p" stands for negative log.\nSo, the negative log of the concentration of H+ ions in solution gives you the pH.  \n-log[H+] = 3.40\nMultiply both sides by negative one:\nlog[H+] = -3.40\nNow raise 10 to the both sides.  This will cancel out the log function. \n10^log[H+] = 10^-3.40\n[H+] = 3.98E-4 M\n\nTo get the [OH-], just divide the [H+] into 1.00E-14:\n1.00E-14 / 3.98E-4 = 2.51E-11 M
165248
(.002mol /20ml) (1000ml / 1L ) = .10 M NaOH\n\n\n pOH =  -log 0.10 =  1\n\n pH = 14.00 - pOH   = 14 - 1  = 13 = D