what is the hydroxide ion concentration of a solution with a pH of 4?
411083
If the pH is 4, the pOH is 10 (pH + pOH = 14).  10^(-pOH) = [OH-].\n\nAnswer is 1E-10
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+].
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
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.
109387
Maybe you are referring to aqueous solutions of these species? If so, then a solution of glucose will be pH 7 since it does not have any proton or hydroxide to change the neutral pH of water. As for solutions of the rubidium, potassium and cesium IONS, they can accept hydroxide from water if they are stable enough to form the metal hydroxide or if they convert to metal oxides. That means hydrogen ions are also formed and this makes the solution acidic, and hence the pH drops. But large cations like Rb and Cs do not readily take in hydroxide owing to the low polarizability of the cations (see me winking here ;p  ), unlike the small K ion. So, with these clues, can you now find out by yourself the answer to your question? Don't we just love chemistry!
518068
Make a solution saturated with calcium hydroxide.  If there is precipitate calcium hydroxide then that means there is as much dissolved calcium hydroxide as possible.  Filter the solution to get rid of the solid.  Now you have a saturated solution, and your objective is to determine the molarity of this solution.  Measure the volume and keep track of this measurement.  To determine the number of moles of Ca(OH)2, titrate it with strong acid like HCl, and see when the pH reaches 7.  When you titrate, keep track of the amound of acid you are adding, and the current pH of the solution.  At pH=7, the number of moles of HCl you added is equal to the 1/2 the number of moles of Ca(OH)2 (because the molarity of H+ and OH- are equal).  Take this number of moles and divide it by the original volume.  This value is the concentration of hydroxide [OH-].  Half of this value is the concentration of calcium [Ca++].  Ksp = [Ca++]·[OH-]²
809057
First of all, basic solutions contain more Hydroxide ions OH- than hydronium ions H3O+ (also known as H+)\n\nAny substances that when dissolved (ionized) in water and produces OH- ions are basic: NaOH becoes Na+ and OH-. NaOH is thus basic\n\nWe know that only weak bases/acids will produce a salt through neutralization which will affect the pH of the solution. For example, the conjugate acid of NH3 is NH4+. NH4+ can react will H2O to form NH3 and H3O+. Therefor, the solution will be more acidic. So we can conclude that positve charged ion can make the solution more acidic and negatively charged ions will make the solution more basic.\n\nHope this helps.
582005
pH=13.36\n\nNaOH is in excess (.0397 mol-.025 mol = .0147 mol)\nDivide by total volume (.0147/(.0397L + .025L) = 2.27x10^-1)\nCalculate pOH = -log (2.27x10^-1) = .6435\nSubtract 14 to find pH 14-.6435 = 13.36
1073643
the easiest way is to use a pH paper to test the pH. and just compare the reading with the one given with the test kit. \n\nOr you may do some neutralisation. And then calculate with the  amount of alkali you use for neutralisation. By that you can calculate the concentration of hydroxide ions in the alkali that you use. Then calculate the pOH for the alkali you use. pH=14 - pOH.\neg. \n1mol of NaOH gives you 1 mole of OH ions.\nso the pOH = -log1 = 0\npH = 14 - 0 = 14\n\nJust some rough guide. Hopefully it helps...Good luck.
592167
We assume NaOH is a strong base and therefore dissociates completely at 60 degrees C. a concentration of .002M solution of NaOH dissociating completely contributes .002 mols/liter of OH ions to the solution. Since pOH = -log[concentration of OH-], pOH = -log.002 = 2.699 (with correct logarithmic sig figs). pH is 14 - pOH, so the pH of the solution will be 14-2.699 = 11.301.
557422
First, let's calculate the pOH of the solution:\npOH = -log [OH-] \nBecause KOH is a strong base (very soluble and complete dissociation), the concentration of KOH is the same as the concentration of the OH- because all of the KOH becomes K+ and OH- ions.  \npOH = -log [OH-] = -log 4.8 x 10^-2 M = 1.32\n\nNow, let's calculate the pH.  \npOH + pH always adds up to 14 under normal conditions.  \n1.32 + pH = 14\npH=12.68\n\nNow, we can calculate the H+ concentration. \npH=12.68=-log[H+]\n12.68 = -log[H+]\n-12.68 = log[H+]\nTake the antilog of both sides, which is the same as ten raised to the both sides. \n10^(-12.68)= 10^(log[H+])\n10^(-12.68)= [H+]\n2.1 x 10^-13 M= [H+]
745984
To answer this question, you must know the Ka or Kb values for those acids or bases above.\nYou must then use those values to find the concentration of H+ (H30+ depending how you were taught it) or OH- ions.\nthese values can then be used to find pH or pOH depending which ion value you have.\nSimply perform the calculation \n\n-log[concentration H+ or OH-]\n\nthis will give you a pH value if you used H+ ions, or a pOH value if you used OH-. To get the pH from a pOH, you must take the value and subtract it from 14. (14-pOH = pH)
754616
NaOH is a strong base that forms hydronium ions (H3O) when introduced to oxacilic acids.  If you know the equation that relates pH to hydronium ion concentration you should be able to calculate the exact change.
704589
Hydronium ions, in an aqueous solution.  H30+
255966
Use the hendeson-hasselbach equation\n\npH = pKa + log([A-]/[HA])\n\nand that you know the toatal concentration of ions\n.2 = [HA] + [A-]\n\nyou know the pKa (found in textbook or given) 4.18\nthe pH is 4.00\n\nso -.19 = log(A-/HA)  therefore .6457 = A-/HA\nand .2 = [HA] = [A-]\n\n... solve simultaneous equation\n\n[HA] = .122 and [A-] = .078\n\ndouble check\n\nK = ([H+][A-])/[HA]\n\nK = 10^-pKa = 6.45E-5\n[H+] = 10^-pH = 1E-4\n...and it does check out
887006
pH= -log([H30+])\n\n\njust put the #'s in the above eqn.\n\nfor pOH, its pOH= -log([OH])\nand for pH given [OH], its pH= 14 - (-log([OH])
101093
I don't even get your question here... \n\npH is a mesurment of the quantity of Hydrogen Molecules that is available in a solution.  The pH of a solution changes depending on the concentration of salt you put in.