Which buffer pair would you use for a pH = 3.00? for pH = 8.75? for a pH = 9.30?
367522
For pH3; Citrate buffer (Sodium citrate/citric acid) \n\nFor pH 8.75; Borate buffer (sodium tetraborate/boric acid), or TRIS [tris(hydroxymethyl)aminomethane]\n\nFor pH9.3; borate buffer, or carbonate buffer (carbonate/bicarbonate)
117085
Buffering, in chemical terms, is a weak acid and its cooresponding salt or a weak base and its cooresponding salt.  The idea is that the weak acid will go to an equilibrium like the following example.  Suppose that I have a weak acid HA and the salt with the anion A-.  The acid will go into equilibrium:\nHA <---> H+ + A-\nWhat will happen is that if you add H+ ions to the buffer solution (acids will add H+ ions), they will be consumed into the reaction to become more HA.  If you add a base (adding OH- ions), it will react away H+ to form water and force more A- to form.  Either way, this buffer system will resist a change in pH.  That's what buffers do. \nBuffers are essential in our body because even though we consume all of these orange juices and sodas, which are both acidic, our body needs to maintain a very precise slightly basic pH, or else we could die.  This is done with CO2 in solution in our bodies.
323054
You sure did pick a really long question for explaining.  And I'll have to argue that all of the pH's do not have correct sig figs but the numbers are right.\n\n(Sorry about the numbers in the tables but this place doesn't seem to allow better tables)\n\nA) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH\n\n*This is a reaction so we have to use moles*\n\nHAc : 25.0mL * (0.100mol/L) * (1L/1000mL)= 0.00250 moles\n\nNaOH : 10.0mL * (0.125mol/L) * (1L/1000mL) = 0.00125 moles\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00125      0.00250        0\nChange:   -0.00125     -0.00125      +0.00125 \nEquilib:      0         0.00125       0.00125\n\nHAc and Ac-(NaAc breaks apart) for an equilibrium\n*Use Molarity when no reaction is taking place (best shown by when no volume is changing)*\n\nBoth HAc and Ac- have same molarity:\n\n(0.00125 moles / 35mL ) * (1000mL/1L) = 0.0357M HAc and Ac-\n\nUsing the equilibrium equation:\n\nEquation:  HAc   <--> Ac-     +  H+         Ka = 1.8 * 10^-5\nInitial:   0.0357M     0.0357M  \nChange:    -x          +x         x\nEquilib:   0.0357-x    0.0357+x    x\n\nKa=1.8*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]\nKa=1.8*10^-5 = [0.00357][x]/[0.0357] usually good approx.\nKa = 1.8 * 10-5 = x = [H+]\n\npH = -log [H+] = -log [1.8*10^-5] = 4.745 (3 sig figs as only after the decimal counts for logs)\n\nB) 25.0mL HAc and 20.0mL NaOH\n\n*See A for HAc moles and double the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00250      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:      0            0          0.00250\n\nCalculate molarity of Ac-:\n\n(0.0250moles /45mL) * (1000mL/1L) = 0.0556M Ac-\n\nUsing the other equilibrium equation (Kb):\n\nEquation:  Ac- + H20   <--> HAc     +  OH-         Initial:   0.0556M   -        0\nChange:    -x        -        +x        + x\nEquilib:   0.0556-x  -       +x         + x\n\nThis gives us the Kb equation\nKb = ([HAc][OH-])/[Ac-] = with approx ([x][x])/[0.0556]\nso Kb = x^2 /0.0556\n\n**but we need to calculate Kb\n**Kw = 1*10^-14 at STP I'll assume this for this problem\n**Kw = KaKb so Kb = Kw/Ka\n**Kb = (1*10^-14)/(1.8*10^-5)= 5.56*10-10 \n\nso x = 5.56 * 10^-6 = [OH-]\n\npOH = -log [OH-] = 5.253\nPH = 14 - pOH (at STP)= 8.747\n\nC)25.0mL HAc and 30.0mL NaOH\n\n*See A for HAc moles and triple the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00375      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:   0.00125            0       0.00250\n\nAfter the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.\n\nCalculate NaOH molarity\n(0.00125moles/55.0mL)*(1000mL/1L)= 0.0227M NaOH\n\nStrong Bases dissociate 100% so [OH-] = 0.0227M\n\npOH= -log[0.0227] = 1.644\npH = 14-pOH = 12.356\n\nThis should show how they calculted the half-neutralization point (A), the neutralization point (B) and after the neutralization point (C).\n\nIf you have questions on this, feel free to send me a message.  I taught this a couple weeks ago so I'm still pretty fresh on the subject.
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?
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
133368
Normal ph level is 7.35-7.45
707264
Methylamine (CH3NH2) and methylammonium chloride (CH3NH3Cl) would be a good buffer in both are in solution together in appreciable concentrations.\n\nNitrates, chlorides, and sulfates (either as acids or salts) will not act as buffers as those anions have no tendency to act as bases (they are the anions from strong acids, which are "strong" because they dissociate completely in solution).  The same with alkali metal hydroxides, which are strong bases.
696123
buffer solution ?\nBuffer solution is  solution of which the pH does not change significantly on the addition of a little acid or base.\n\nBuffer are also important in many industrial processes.Certain processed food has to be prepared in a buffered form so that when consumed does not cuase undue changes in pH.
451857
Chemicals used in the reaction is important. What is the pKa of the reactant? Research on which buffer and what its pH buffering region is important, because it should cover the pKa region of the chemicals used. Once knowing which buffer to make, we can take a buffer by making a mixture that have a weak acid with its salt of conjugate base, or weak base with its conjugate acid. It should be noted that here buffer solutions are presented in terms of the Brønsted-Lowry notion of acids and bases.
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.
894546
You did not put the volume of HCl to neutralize so we will do the neutralization to pH = 7 only.\n\nmoles of HCl = 35 0 g/l / 36.46 g/mole = 9.60 mol/l\nmoles of NaOH = 30 g/l / 40 g/mol = 0.75 mol/l\nOne mole of HCl will neutralize one mole of NaOH.\n\n9.60 * Vol(HCl) = 0.75 * V(NaOH).\n\nOnce you know if you have any NaOH left you can calculate how much NaOH to add to bring it to pH = 8 or 9\n\npH = 8 =====> pOH = 6  (pOH = 14 - pH)\nthen 6 = - log [OH]\nCalculate OH from last equation and this will give you the moles of NaOH you need to add to bring to pH = 8.  Same for pH = 9.
858181
It would be 1.8N.  \n\nThree times in the case of Phosphoric Acid.
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.
1088951
Carbonic Acid is an example of a buffer. It keeps the equilibrium in your blood stream, so that the pH remains around 7.2-7.4. If there are too much acid, it acts as a base to reduce the amounts of acid, and vice versa. \n\nSuppose you injected yourself with sodium hydroxide (why, I don't know), then:\nNaOH + H2CO3 = NaHCO3 + H2O \nBut if you drank HCl instead of H2O:\nHCl + NaHCO3 = NaCl + H2CO3 \n\nObviously, the pH in your mouth is different from that of your stomach, seeing as how stomach acid is HCl. but naturally, without additional acids and bases, they are all roughly the same.
255966
Use the hendeson-hasselbach equation\n\npH = pKa + log([A-]/[HA])\n\nand that you know the toatal concentration of ions\n.2 = [HA] + [A-]\n\nyou know the pKa (found in textbook or given) 4.18\nthe pH is 4.00\n\nso -.19 = log(A-/HA)  therefore .6457 = A-/HA\nand .2 = [HA] = [A-]\n\n... solve simultaneous equation\n\n[HA] = .122 and [A-] = .078\n\ndouble check\n\nK = ([H+][A-])/[HA]\n\nK = 10^-pKa = 6.45E-5\n[H+] = 10^-pH = 1E-4\n...and it does check out
952799
"p" is defined as minus the log...pH is saying -log(H).\n\nTherefore, -log(4) = 1 x 10 to the minus 4 = .0001\n\na pH of 8 = 1 x 10 to the minus 8 = .00000001\n\nso the change in concentration is 10 to the 4 or 10,000 times
769229
At the equivalence point the curve changes its curvature. Thus instead of trying to find the middle of the curve in order to determine the endpoint you can find it by doing the 1st derivative (the endpoint will be a peak) or the 2nd (at the endpoint it will be 0))\nSince you are using experimental data, when you are calculating the derivatives you will have an error depending on the amount of base added at each step (the bigger the volumes, the bigger the error; remember that detivatives require by definition extremely small changes in order to be accurate). I guess your instructor told you to take the average of the two in order to minimize the error.\n\nThe equation is actually pH=pKa+log[conj.base]/[acid]. At half equlivalence point  [conj.base]/[acid]=1 thus pH=pKa+log1=pKa.\n\nIf your equivalence point is at 6.97 ml then the half equivalence point is at 3.48-3.49 ml. Go to your pH vs V base plot, and find the pH value for that volume. For that value pH=pKa and thus Ka=10^-pKa