when titrating a weak acid with a strong base, the pH is not equal to 7.00 at the equivalence point. Why?
357070
The equivalence point is where all the reactants have reacted.  Since you have a strong base and a weak acid, the pH of the base pulls it over to the basic side of the scale rather than neutral.  Simply, there are more OH- ions than H+ ions total.  All the H+ ions are reacted so you are left with extra OH- ions.  Also works vice-versa.
452992
Well first of all you have not given us all the information to be able to answer this buffer question.  We need to know the Ka (6.2x10-8)of the acid.  But the steps to perform the task.  Calculate how much of the acid is neutralized by the addition of the base.  Recalculate the equilibrium concentration of both the acid and base with the new acid value.  using Ka=[H+][Na2HPO4]/[NaH2PO4] solve for H+ and finally use pH=-log[H+].  Your final answer..pH=7.25
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+
529279
Acids.\n\n0-7 = Acid\n7 = Neutral\n7+ = Base
716967
It has to do with the definition of pH, which is negative one times the base ten log of the concentration of H+ ions in a solution.  The dissociation constant of water happens to be 1.011 x 10^-14, yielding a pH of 6.99, or experimentally 7.  It is essentially a coincidence.  If the dissociation constant of water hadn't been near a power of ten, there would probably be a normalization constant involved, and the scale could be centered on zero or any other arbitrary value.
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.
117935
I believe that it is 7... If the concentration of the acid and base are equal then they would probably neutralize the solution.
936672
The pH level. 7 is neutral. Below 7 is acidic and above 7 is basic. 1 being most acidic and 14 being most basic.
378003
pH for a salt of strong base and weak acid is given by:\n         pH = 7 - (1/2)pKb - (1/2)log C\nwhere pKb = dissociation const of strong base\n       C  = concentration of Acid.\n\nKb = Kw / Ka = (10^-14) / (6.5 x 10^-5) = 1.53x10^-10\npKb = -log (1.53x10^-10) = 9.8154\nlog C = log 10^-1 = -1\n\npH = 7 - (1/2)x 9.8154 - (1/2)x-1\n   = 7 - 4.9077 + 0.5\n   = 2.5923
282642
actually, pure water does not have a pH of 7.0, it is slightly acidic, around 5.6-5.8 from what i have seen at work.
234325
what the previous writer has posted is true only for sodium. equivalence takes into account both the molecular or atomic weight of the species you're interested in and its valence - for example sodium has atomic weight of 23 and a valence of 1 (the ion is +1), hence the 23/1 conversion in the previous answer. This wouldn't be the same for, say, carbonate with mw ~60 and valence of 2 (= equivalence of 30)
801550
You have\nmole CH3COOH= Ca*Va= 0.2*0.2=0.04\nmole KOH= Cb*Vb= 0.2*0.2= 0.04\n\nCH3COOH + KOH -> CH3COOK + H2O\n\nThe stoichiometry is 1:1 and you have equal mole of acid and base therefore they will react completely and you will have only CH3COOK in the final solution\nHowever this is the salt of a weak acid with a strong base so it will hydrolyze according to\n\n.. .. .. ..  .. CH3COO- + H2O <=> CH3COOH + OH-\nInitial .. .. .. .. C\nReact .. .. .. ..x\nProduce .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. . x\nAt Equil. .. .. C-x .. .. .. .. .. .. .. .. .. .. .. x .. .. .. . .. .. x\n\nKb=[CH3COOH][OH-] / [CH3COO-] = x^2/(C-x)\n\nBut Kb=Kw/Ka= 10^-14/(1.82*10^-5)\nC=moleCH3COOK/Vfinal = Ca*Va/(Va+Vb)=0.04/0.4\nsubstitute Kb,C and find x (you can assume that C>>x and C-x=C. this simplifies the equation to x^2/C=Kb. If you calculate x and it is truly much smaller than C the assumption is valid. Otherwise you have to solve the quadratic equation)\n\npOH=-logx and pH=14-pOH
880704
The pH of the salt depends on the strengths of the original acids and bases:\n\nAcidBaseSalt pH\nstrongstrongpH = 7\nweakstrongpH > 7\nstrongweakpH < 7\nweakweakdepends on which is stronger\n\nThat should be what you want.
925984
YES, EG. SALIVA FOR EXAMPLE IS MORE BASIC THAN ACIDIC, AS IS ALSO  HUMAN BLOOD. THE Ph OF BLOOD IS 7.62,  NEUTRAL Ph IS 7.0.
211023
wow, well put mrjeffy! I have one thing to add though. Because all acids and bases are electrolytes, you can use pH to determine what is and isn't an electrolyte in a solution. If it come out as acidic or basic, it is and electrolyte(the darker, the more basic or acidic). If it comes out neutral, like Distilled Water, it is not. Also don't forget that non-electrolytes will not completely dissolve into solution in water.
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?
556259
strong acids have a high tendency to donate protons (hydrogen ions) and strong bases have a high tendency to receive them.  \nstrong acids have a very low pH, 1-3\nweak acids have a ph of about 3-6\npH 7 - neutral solution\nweak bases have phs of about 8-11\nstrong bases - ph 11-14