What is the pH of a 0.00339 M solution of NaF, given the Ka of HF= 6.8x10^-4 at 25 degrees C?
438792
Okay. This is a multi-step pH problem. First, find the Kb of the fluoride ion (F-). To do this, follow this formula:\n\nKw/Ka = Kb\n\nThis formula can be rearranged to be Kw/Kb = Ka, by the way. Kw is the K of dissociation of water, which is 1.0e-14.\n\nPlug in the numbers and you get 1.47e-11 as a Kb for fluoride. Now, write an equilibrium ICE table, filling in the values you know.\n\n_F- + H2O --> HF + OH-\nI\nC\nE\n\nSolve the equilbrium table and you get the following equation:\n\nx^2/(0.00339-x) = 1.47e-11\n\nx is the concentration of both HF and OH-.\n\nSolve the equation and you get 2.23e-7 as the concentration of OH-. The easiest way to finish the problem now is to find the pOH. Take the -log of the concentration of OH-. The pOH is 6.651. Now, follow this equation: pH + pOH = 14. The pH of the solution is 7.34876. To the right amount of significant figures, the answer is 7.349 (the numbers to the left of the decimal place doesn't count toward the sig figs in pH's).\n\nHope this helped.
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.
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.
225870
Take 33g of the salt and dissolve it in 67 g of water to get 33% salt solution.\n\nIts not the same as 0.3 N solution
464163
FORMATION CONSTANT IS :\n\nkf = 1.5*10^7
232172
ΔT = 80.6 °C - 75.4 °C = 5.2 °C; m =  ΔT/ Kfp = 5.2 °C/(6.9 °C/m) = 0.75 m; m = moles solute / 1000 g solvent, so that 1.00 g/ 12.5 g = x g/1000 g, x = 80 g, and 80 g = 0.75 mol, or molar mass = 80 g/0.75 mol = 1.1×10^2 g/mol.\n\nHope this helps,\nUltimeaciax
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
657714
NaOH + HCl --> NaCl + H20\n\nn(NaOH) = 1.00 x 32.6cm3\n        = 1.00 x 32.6 x 10^-6\n        = 3.26 x 10^-5  mol\n\nn(HCl) = 1.00 x 25cm3\n       = 1.00 x 25 x 10^-6\n       = 2.5 x 10^-5  mol\n\ntherefore, NaOH is in excess.. hence the final solution is alkaline, which means the pH is more than 7.\n\nDue to the limitation of my knowledge, i can't give you the exact pH value. Just hope that it will help you to continue solving...
968497
The initial concentration of the acid is C=mole/V=1/2=0.5 mole/L\nLet's assume that the acid is HA\n\n.. .. .. .. .. .. .. HA <=> H(+) + A(-)\nInitial .. .. .. .. 0.5\nDissociate .. .x\nProduce .. .. .. .. .. .. .. x .. .. .. x\nAt Equil. .. . 0.5-x .. .. . x .. .. .. x\n\nKa=[H+][A-]/[HA] =x^2/(0.5-x)\n\nbut x=[H+]=3.4*10^-2 so\n\nKa= ((3.4*10^-2)^2) / (0.5-3.4*10^-2) =2.48*10^-3
819244
a) Acid +H2O -> H3O(+)+ Basis\n\nKa1 =  x²/(0.1-x) = 10-4\nx<< 0.1 so x = 10-3\n\nso pH = 3\n\n2) second dissociation\nKa2 = y (10^-3 + y) / (10^-3 -y)\ny << 10¨-3\ny = Ka2 = 10^-8
243948
Yes, using the David Hasselhoff equation is much easier. The HH equation is:\n\npH = pKa + log(base/acid)\n\nFill in the information you know (either from the problem or a table).\n\n9.00 = 9.25 + log(0.10/x)\n\nSolve for the concentration of NH4.\n\n0.178 = [NH4+]\n\nSince there is 2.0L of solution, use that to solve for the number of moles.\n\n0.178M * 2.0L = 0.356 moles NH4Cl.\n\n\n\nIn case you were wondering, you have to look up the Kb of ammonia (NH3) and convert it to the Ka of ammonium.\n\nKw/Kb = Ka\n\n1e-14/1.78e-5 = Ka = 5.62e-10\n\nTake the -log of that to find the pKa of ammonium, which is needed in the equation.\n\n\nHope this helped.
327703
The amount is diluted by a factor of 25.  So the resulting solution would have a concentration of 11.6M divided by 25 or \nor 0.464M.
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?
968707
NH3  -33.34 °C\nCH3F -78°C\nCH4 −161.6 °C\nHe -268.93 °C
736124
The answer to the first part:\n1) You need to find the pKa ---> use pKa+pKb = 14\nso 14-pKb = pKa\n14-4.75 = 9.25\n\n2) work out the final Molarity of each\n[NH3] = 0.400M/0.480L = 0.833M\n[NH4Cl] = 0.100M/0.520L = 0.192M\n\n3) now use the henderson-hasselbach equation\npH = pKa + log([base]/[acid])\n= 9.25 + log([0.833]/[0.192])\n= 9.89
173742
To determine pH for this mixture, one must know the molarity for the HCl, the molarity for the acetic acid, and the Ka for the acetic acid.  The goal is to find the concentration of the hydronium ion in solution.  Thus, for a strong acid, since it completely dissociates, the molarity equals the concentration of the hydronium ion.  But, it is more complicated for the acetic acid, which is weak.  We need to solve for the hydronium concentration, so we set up a Ka equilibrium equation, which with all the values added looks like:  Ka= (x^2)/ (Molarity of Acid).  Solve for x.  This gives you the hydronium ion concentration for the weak acid.  Add the two concentrations together and take the negative log of that number.\nRemember to use correct significant figures.  Also, it is not necessary to subtract x from the molarity of the acid in the equation, because since the Ka is small enough, when the math is done the percent ionization is less than 5%, so the x can be ignored without incurring significant errors.
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)