What is the pH of a 2.65 x 10^-5 M solution of HClO4?
931053
Perchloric acid is one of the six common strong acids, which means 100% dissociation in water.  Thus, [HClO4] = [H+].\n\npH = -log 2.65 x 10^-5 = 4.58
628177
HCL   +   H2O     ->   H2O     +    H+    +   CL-\n1X10-4 M  1X10-4M  ->  1X10-4M    1X10-4M     1X10-4M\nPh = -log (1x10-4)\n   = 4\nPoh=14-4\n   =10\n(OH-) conc. in Mol/dm3 = 1x10-10 M
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.
875212
Since HCl and NaOH are both strong solutions (acid and base, respectively), you know the pH is going to be 7.0.
1108880
HCl   +  NaOH   ===   NaCl   +   H2O\n0.0015 ,  0.0016 ,             0        ,       0        at start of reaction (moles)\n0,           0.0001 ,            0.0015 ,    0.0015  at end of reaction(moles)\n\nevery mole of water contributes about 10^-7(ten to the power -7)\nmoles of OH-  and H+ . thus their contribution to the pH is neglected when there is a considerable amount of acid or base present  i.e more than 10^-5 moles/litre (not moles but CONCENTRATION)\nThus concentration of NaOH = 0.0001/90*1000 moles/litre{total volume = 50+40 ml} = 0.001 .\nas NaOH is a strong electrolyte like any other sodium compound  we can assume complete ionisation . thus[OH-]=[NaOH] = 0.001. now pH = -log[H+]=14-(-log[OH-])= 14+log[OH-]  =14+log[0.001]=11\nthus the pH=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
411083
If the pH is 4, the pOH is 10 (pH + pOH = 14).  10^(-pOH) = [OH-].\n\nAnswer is 1E-10
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
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])
844919
4.27 x 10^-3 g NaH2PO4\n\nthe molar mass of NaH2PO4 can be found by using the masses from the elements on the periodic table\n23 + 2(1) + 31 + 4(16) = 120 g/mol\nthis value is used to convert\n\n4.27 x 10^-3 g NaH2PO4 x (mol / 120 g) = 3.56 x 10^-5 moles NaH2PO4
646163
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
621533
NaOH ----------------> Na+   +   OH-\nFind the number of moles of NaOH\nMoles of NaOH = 0.10M * 0.150L = 0.015 moles\nSince the coefficients in fornt of NaOH, Na+ and OH- are all 1, therefore if 0.015 moles of NaOH disssociates, 0.015 moles of Na+ and 0.015 moles of OH- are produced.\n\nSimilarly we will work with HCl\nHCl --------------------> H+      +      Cl-\nMoles of HCL = 0.30M * 0.05L = 0.015 moles\nThis means the 0.015 moles of H+ will be produced\n\nSince H+ and OH- have equal amiunts, they will completely neutralize each other leaving the solution at neutral pH. Since the neutral pH is 7, therefore, the pH of this solution will be 7.
465978
45 cm3 = 45 mL\n\n45 mL * 1.2 M HCl = 54 mmol HCl\n\n60 mL * .9M HAc = 54 mmol HAc\n\nSince the material amount is the same, HAc (acetic acid, CH3COOH) will have a higher pH because HCl is a strong acid and it will have a low pH. CH3COOH is a weaker acid, so its pH would be higher. HCl is a strong acid because the H and the Cl are strongly solvated by water molecules because of the immense polarity of the H-Cl bond. CH3COOH is weak because, while the anion is resonance stabilized, you still have a molecule with a hydrophobic end and an electron donating group. (the CH3)
490327
solution A has a hydrogen ion concentration of 10^-5\nsolution B has a hydrogen ion concentration of 10^-3\nAll you have to do is 10^-ph\n\nIs this homework?
318573
The reaction between Hydrochloric acid and Sodium Hydroxide is as follows,\n\nHCl + NaOH --> NaCl + H2O\n\nYou can see that there is a 1:1 ration between the number of moles of HCl needed to completely reacted with NaOH.\n\nSince you are given both a volume and concentration of the NaOH solution, you should be easily able to calculate the number of moles of NaOH availible.\n# mole = volume * concentration\n# moles = .1 Liters * .25 moles/L = .025 moles of NaOH\n\nIf you know you have .025 moles of NaOH, you know you need .025 moles of HCl to react it completely.\n\nIf you know that you have a .15 mole/Liter solution of HCl, you can calculate the volume required to get .025 moles lie this,\nLiters = moles / oncentration\nLiters = .025 moles / .15 moles/Liter = .01667 Liters or about 167 mL of solution.\n\n\nNext time do your own homework.
950979
pH = - log[H+], so [H+] = 10^(-pH)\n\nfor pH = 7, [H+] = 10^(-7)
371296
well seeing that the molarity of the two solution is a direct ratio of how many molecules there are in the solution, i think that the ration would be 0.15/0.20\n\nso assuming that the reation is 1 to 1 (i cant remember so if it isnt 1 to 1, then i'm wrong) then you would need 18.75 mL of the 0.200 M HCl