what general pH for normal lake?
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+].
1171925
PH of the aqueous solution of NaCl is 6 - 7 @ 1% by weight
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.
48939
none water is H2O two parts hydrogen 1 part oxygen any thing else the water is inpure.but in the case of fish do a 1/4 water change every week so not to shock your fish make sure you keep temp about 78 degrees or they can get ick as far as testing you dont have to test fresh water but salt water is where the nitrite cycle is critical and then you also will test ph which is idealy 8 use a hydrometer to the line as it mesures the salt level by the boyency of the device nitites no higer than 14 i belive ask you pet shop owener as it been abuot 10 years since i had my salt water tank but for fresh water, the water change should cover it.
410035
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.
440505
pH = pKa + log ([Salt]/[Acid])\n\n[Salt] = 0.5 ; [Acid] = 0.75 ; ([Salt]/[Acid]) = 2/3\n\npKa = -log (1.85 x 10^-5) = 5-0.2671\n\nlog ([Salt]/[Acid]) = log 2/3 = -0.1761\n\nHence, pH = 5 - 0.2671 + (-0.1761)\n          = 5 - 0.4432\n          = 4.5568\n\nNote: All Calculations done using Clark's Tables. Use your calculator to calculate log for greater accuracy.
271476
pH = 4.75\n\ntechnically, the pH can be anything.  But, for the most effective acetic acid buffer, the pH would be 4.75 because the pKa of acetic acid is 4.75.  When the pH = pKa, the amount of acid and base are equal, thus its resistant to pH changes is greatest.
209635
Go out to the lake and stick your finger in it. Or buy a fish finder with temp on it and check it out for yourself. I know of nowhere that you can get water temp other then I have heard them give it on lake Michigan. From buoys I think.
75905
From what I can find out, it appears to be slightly acid, approaching neutral, pH 5-7\n\nHow cool is that?! It's like that question was designed just for someone with my name!
122833
pH level, buffering capacity (alkalinity), general hardness and salinity
455471
pH (potential hydrogen) is a measure of the activity of hydrogen ions (H+) in a solution and, therefore, its acidity or alkalinity. In aqueous systems, the hydrogen ion activity is dictated by the dissociation constant of water (Kw = 1.011 × 10−14 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\npH = -log(concentration of H+)\njust use the equation to figure it out.\nhttp://en.wikipedia.org/wiki/PH
249301
H3O+ .0292M = pH 1.53
1186868
Sodium Hypochlorite can exist at any pH that is on the "basic" side of neutral. Having said that, most commercial sodium hypochlorite solution (read "Liquid Pool Chlorine") is stabilised by the addition of sodium hydroxide and so has a pH between 11 and 12.
171909
IgDs that naturally have a neutral to alkaline ph
538874
I think both your answers are right. \n\nAccording to my books, normal values are pH = 7.4, PCO2 = 40mm Hg, and HCO3 = 24 mM.\n\nRespiratory acidosis: pH < 7.4, CO2 too high\nMetabolic acidosis: pH < 7.4, HCO3 too low\n\nRespiratory alkalosis: pH > 7.4, CO2 too low\nMetabolic alkalosis: pH > 7.4, HCO3 too high\n\nSo the way I figure out these problems is first to look at the pH. Even if compensated, it will never get back to normal so it should tell you right away if it's acidosis vs alkalosis. \n\nThen look at both the PCO2 and HCO3. Are they both above normal, below normal, or is one above and the other below? If one is above and the other below, then it means that the state is uncompensated and it should be easy to figure out the cause.\n\nIf they are both either below or above normal, then you have to look at either CO2 or HCO3 more closely. I usually like to concentrate on HCO3, but it really doesn't matter which one you want to look at. \n\nSo looking at your problem 1 for example, pH - 7.63 tells me its alkalosis. Both CO2 and HCO3 levels are lower than normal, so I know it's being compensated for. From what I know about alkalosis, it's either caused by low C02 or high HCO3. Since HCO3 is low in this case yet the blood pH is basic, I know it can't be the cause of the alkalosis. CO2 however looks to fit into the model, so HCO3 must be compensating.\n\nAnyway, hope this makes sense, kinda hard to explain completely in a post.
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.
241965
Check out the buffer recipe calculator at <http://www.bioinformatics.org/JaMBW//5/4/index.html>\n\nSome things to consider:\n\nAt what temperature do you want the pH to be 7.5? (the pKa of tris-base varies as a function of T)\n\nWhat ionic strength do you need?