What is pH when .5 moles NaHCOO is dissolved in 1 liter of .5 M HCOOH? (Ka=1.8 times 10 to th -4th for HCOOH?
397741
by the simplified version of the Henderson hasselbach equation\n\npH=pka + log(A-/HA)\n\npH=-log(Ka) + log (1) = -log(Ka) + 0 = -log(1.8x10^-4)= 3.77\n\n- 2 sigfigs! Since when you take the log you should have as many decimal places as numbers you started with.\n\n-By consequence, this is at the point known as the Equivalence point.
368556
The pH of a solution tells you the concentration of the H+ ions in solution.  However, if we know the pH, we can calculate the pOH and from that, the concentration of the OH- ions in solution.\n\npH + pOH = 14\nif pH = 8.25, pOH = 14-8.25 = 5.75\n\nJust like pH = -log([H+]),\npOH = -log([OH-])\n\nIf pOH = 5.75, that means that -log([OH-]) = 5.75, which mean that 10^-5.75 much give us the concentration of OH- ions.\n\n10^-5.75 = 1.778 E-6\nSo we know that the concentration of the OH- ions in the solution with a pH of 8.25 is 1.778 E-6 Molar.\n\n[OH-] = 1.778 E-6 M
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.
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
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?
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.
971190
Molarity is defined as number of moles in 1 liter.\nYou have 0.01 mol and 10 liters\nM = 0.01 / 10 mol/liter - 0.001 mol/liter = 0.001 M.  That's it.  You did it.\n\nNow pH = - log[H+].\n\nIf you write the equation\nHCl ---> H+ + Cl-\nThis means that you have the same molarity for HCl and H(+).  For both of them the coefficient in the chemical equation is equal to 1.  For other problems the coefficients may be different so always look at the balanced equation.\n\nso M(HCl) = M(H+) = 0.001.\n\nPutting this value into the pH equation:\n\npH = - log 0.001 = 3\n\npH + pOH = 14.\nCalculate the pH according to the pH definition and then use the above equation to find the pOH.
780228
HCL and NaOH are 'strong' acid and base. When they react, the product is neutral. NaOH + HCL --> NaCl + H2O\n\nWhat you need to find is which is the limiting reactant, the other one (what is left over) will effect the pH by its concentration in the otherwise neutral solution.\n\nFor your first problem:\n\n0.0557M HCL = 0.0557mols/liter*0.00542L = 3.02*10^(-4)mol HCL\n\n0.0377M NaOH = 0.0377mols/liter*0.00337L = 1.27*10^(-4)mol NaOH\n\n3.02*10^(-4)mol HCL - 1.27*10^(-4)mol NaOH = 1.75*10^(-4)mols HCL\n\n0.00542L + 0.00337L = .00879L\n\n[(1.75*10^(-4)mols HCL) * (1 L) ] / (.00879L) = .020M HCL\n\npH = -log(.02) = 1.70
553835
Well to determine [H+] you use the equation that \npH=-log [H+] \nthus a solution with pH=3 will have a [H+] of 1x10^-3, \nand a solution with pH=8 will have a [H+] of 1x10^-8\nTo find [OH-] there are several equations you can use\n[OH-]=Kw/[H+], \nKw is the ionization constant of water which is equal to 1.0x10^-14\nThus you will have [OH-] of 1.0x10^-11 and 1.0x10^-6\nSo you answer will be that the solution with pH=3 will have a [H+]>[OH-]
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
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+
411083
If the pH is 4, the pOH is 10 (pH + pOH = 14).  10^(-pOH) = [OH-].\n\nAnswer is 1E-10
802376
YOUR ANSWER IS CORRECT!  Here's how i did it:\n\nYou can find the # mol of NaOH used:\n\nMolarity = (# mol)/(volume solution)\n0.4 = (# mol)/(.075 L)\n\nSolving, you get 0.03 mol of NaOH.\n\nUsing the balanced equation:\n\n3 NaOH + H3PO4 ==> Na3PO4 + 3 H2O \n\nYou see that 3 mol NaOH = 1 mol H3PO4.  Thus, if we have 0.03 mol NaOH, that means we have 0.01 mol H3PO4.\n\nNow we can solve for the molarity:\n\nMolarity = (0.01 mol H3PO4)/(0.05 L) = 0.2 M
633715
5. KHCrO4 ----------------> K+   +   HCrO4-\nSince KHCrO4 dissociates complteley,\n the molarity of HCrO4- = molarity of KHCrO4\n the molarity of HCrO4- = 0.05M\n\nNow HCrO4 ------------------>    H+    +     (CrO4)2-\nCalculate the concentration of H+ from the pH value\n[H+] = 10^(-pH)   = 10^(-3.80)  = 1.58 * 10^(-4)\nIt is clear the only source of H+ is HCrO4-, therefore the concentration of (CrO4)2- is the same as concentration of H+\n[(CrO4)2-]   =   [H+]   = 1.58 * 10^-4\n\nKa = {H+} * [(CrO4)2-] / [HCrO4-]\nKa = [1.58 * 10^(-4)]^2/0.05\nKa = 5.02 * 10^ (-7)\n\n6.       KOH----------------> K+    +    OH-\n         HClO4 -------------> H+    +    ClO4-\n(a) \n    Moles of KOH = Molarity * Volume = 0.200M * 0.030L\n    Moles of KOH = 0.006 moles\n    From the equation above, \n    Moles of OH- = moles of KOH = 0.006 moles\n    \n    Moles of HClO4 = Molarity * Volume = 0.150M * 0.030L\n    Moles of HClO4 = 0.0045 moles\n    From the equation above,\n    Moles of H+ = moles of HClO4  = 0.0045 moles\n\n     H+    +    OH-    ------------>   H2O\nWhen this reaction is finished, 0.0015 moles of OH- are left\n[OH-]  =  moles of remaining OH- / Total Volume\n[OH-]  =  0.0015 moles/ 0.06L (30ml + 30ml)\n[OH-]  =  0.025M\npOH = -log [OH-]\npOH = -log (0.025) = 1.60\npH = 14 - pOH  = 14 - 1.60 = 12.40 \n\nDo the same for parts b, c, d, and e\n\n7. I will assume that \n   H2S --------------------------> 2H+    +    S2-\n   Ksp of HgS = 2.0 * 10^(-53)\n   Ksp of FeS = 6.3 * 10^(-18)\n\nKsp of HgS = [Hg2+] * [S2-]\n[Hg2+] = 0.01 M\n[S2-] = x\n2.0 * 10^(-53) = x*0.01\nx = 2.0 * 10^(-51)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 2.0*10^(-51) = 4*10^(-51)\npH = -log (4*10^(-51)) = 50.39\n\nKsp of FeS = [Fe2+] * [S2-]\n[Fe2+] = 0.01M\n[S2-] = x\n6.3* 10^(-18) = x* 0.01\nx = 6.3 * 10^(-18)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 6.3 * 10^(-18) = 1.26*10^(-17)\npH = -log (1.26*10^(-17)) = 16.90\n\nthe pH range is from 16.90-50.39\nThis is the best explanation I can gibe. By the way, pH level beyond 0 and 14 do exist, but, I'm not 100% about this answer.\nI fixed it, I think it is correct now.
119509
my question is:\n\nNaHCO3(s) + NaOH(aq) -> Na2CO3(aq) + H2O(l)\n\n3.25 moles of Na2CO3 is produces.
386784
Well first of all you have to specify if the H's you want to account for are either comming from HCl alone or both HCl and some H's comming from the dissociation of water.\n\nI suppose that only the number of moles of H's comming from HCl is really worth while working the question for, since the H's of the dissociated water does not change the pH of the 0.10M solution.\n\nTherefore:\n\n0.10moles are in 1000mL\n? moles are in 50mL\n\n50 x 0.1 / 1000 = 0.005moles\n\nTherefore we can say that 0.005moles of HCl in 50mL of water results in a concentration of 0.10M
774733
The number of moles of acid used for neutralisation = volume (in Litres) x concentration  =  0.045 x 0.5  =  0.0225 moles.\n\nFrom the equation for the reaction it can be seen that the same number of moles of NaOH are used as the acid.\n\nNaOH  +  HCl  =  NaCl  +  2H2O\n\nSo there must also be 0.0225 moles of NaOH\n\nThe mass of NaOH  =  number of moles  x  formula mass (Na : 23 +  O  : 16  +  H  : 1) = 0.0225 x 40   =  0.9g NaOH\n\nSo the % NaOH in the Lye is 0.9/1.2  x 100   =  75%