How do you calculate the pH of a Buffer system when NaOH is added?
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
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.
511319
NaOH is in excess\n\nSubtract moles of H+ from OH- (.0345 mol OH- - .025 mol H+ = .0095 mol OH-)\n\nDivide by total volume (.0095 mol OH-/.0595 L = 0.1596 M)\n\nFind pOH = -log [OH-] = -log [.1596] = 0.7967\n\nFind pH = 14-.7967 = 13.20
904575
pH=-log(H+)\nthen H+ or H3O+ if done in water is:\n10^(-1*pH)\nFor plugging in 7.5 in above returns: 3.16 x 10 -8 M (molar concentration)\nsimilarly pH 9.6= 2.51 x 10 -10 M
122833
pH level, buffering capacity (alkalinity), general hardness and salinity
564023
acid + base ---> salt + H2O
904431
Just the take the [H+] ion concentration divided by the initial [HF]:\nfrom your info i gather it should be (10^-1.59)/1.0 \n\nConvert it into a percentage.
864626
1) Calculate the initial concentration of the H+.\npH = - log [H+] = 2, therefore  [H+] = 10 ^ -2 mol / liter\n\n2) Calculate the dilution factor.\nWe want to change the concentration from 10 ^ -2 mol / liter to\n10 ^ -4 mol / liter. Therefore the dilution factor is 100.\n\n3) Calculate the amount of water to add.\nThe dilution factor is 100. We therefore want to have 100 x 14.5 ml = 1450 ml after dilution. We therefore have to add 1450 - 14.5 = 1435 ml of water
950979
pH = - log[H+], so [H+] = 10^(-pH)\n\nfor pH = 7, [H+] = 10^(-7)
568277
Yeah, it can be done with just the line.  \nFirst, locate the end point on the line.  It's exactly the midpoint of the sharp pH rise or drop.  Next, find exactly half the end point, where the moles of salt equal the moles of whatever you're adding.  At that point, pKa = pH.
776525
Since you are pretty, here's how:\n\n1. Compute for the number of moles NaOH needed to prepare the solution.\nmole NaOH: 1 L * 0.5 mol/L = 0.5 mol\n\n2. Get the weight NaOH you need by multiplying the number of moles to the molecular mass of NaOH.\nweight NaOH: 0.5 mol * 40g/mol = 20g\n\n3. To prepare, weigh 20 grams of NaOH pellets. Dilute this with water to 1000 mL in a volumetric flask.
720236
1) Bi carbonate buffer system will contain HCO3- ion\n\nwhen a base is added it will liberate the H+ and produce water combining with OH- ion of the base there for reducing the effect of the base\n\nand when an acid is added it will accept the H+ ion and for H2CO3 therefor the effect of the acid will be nullified!!!\n\n2)phosphates are H2O4- or HPO42-\n\nand it will also have the functions as mentioned in the above!!\n\n3)proteins have an NH2 group and well as a COOH group both attached to the same molecule!!!\n\ntherefor when acid is added the NH2 group will neutralize it\n\nand when a base is added COOH group will neutralize it  reducing the alkalinity!\n\nthere for all these chemicals found in the blood and other body tissue will maintain the blood pH level at a constant level acting as natural buffers!!\n\nhope you understood!!!!!\n\ncheers!!
746933
You could use the Van't Hoff equations to calculate this.\nOP = i * c * RT\nwhere i is the number of ions per particle and the membrane will be retaining the ions.\n\nBecause RT will be constant for all of them, you only need to compare the product i *c for each solution.  So OP/RT for the solutions above will be.\n\nNa2S ===> 0.30 * 3 = 0.9\nMgCl2 ===> 0.2 * 3 = 0.6\nEtOH ===> 0.50 * 1 = 0.5\nKCl ===> 0.50 * 2 = 1.0\nso the answer is KCl, Na2S, MgCl2, EtOH.
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.
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
89429
That would depend on what the system is, what the current pH is, what you are trying to accomplish (create a certain buffer, etc.)\n\nIf the current pH is > 7.5 you would use an acid (HCl, HNO3, etc.) to lower the pH.  If the current pH is < 7.5 you would use a base (NaOH, NH4OH) to raise the pH.
875212
Since HCl and NaOH are both strong solutions (acid and base, respectively), you know the pH is going to be 7.0.