How could you quickly make 100ml of a sodium acetate buffer? What pH should the buffer be made in?
140250
Add 50mL of acetic acid to 50mL sodium acetate. The pH is the pKa, unless unequal amounts of reactants are used.
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.
245193
Use your buffering equation and plug in the numbers you know to get your answer.  You might have to do reverse calulations to find your answer.
334010
The only way I've heard of to obtain pure sodium metal is by electrolysis of NaCl.  Basically, place the salt solution between two electrodes (Na+ attracted to anode, Cl- attracted to cathode).\n\nPure Na is extremely reactive, though.  It is very difficult to store, as any exposure to air will cause almost instant oxidation of the surface (formation of Na2O).  There is also very little commercial use for pure Na.  There are many, many commercially viable sodium compounds, but pure sodium is rarely useful.
83196
I've usually made up a concentrated stock of various buffers and dilute it to the concentration I need.  From a 1M Tris stock it's a 1/100 dilution to 10mM and from a 1M MgCLs it's a 1/1000 dilution.  I would decide how many milliters I wanted to make, measure out the cesium powder then pipette out the amount of each Tris (1/100 of the total solution vulume) and Mg (1/1000 of the volume) then bring to the final volume with distilled water.  This works fine if the concentrated solutions don't interact and cause precipitates or something else like being nearly a saturated solution.
320061
Someone else has asked this same question before, and it really isn't that hard . . .\n\n60 ml of 0.250M HCl plus 60 ml of 0.125M NaOH will result in 120 ml of solution total.  All of the NaOH will have been neutralized by the HCl, leaving an excess of HCl, so the solution will be acidic -- the pH will be much less than 7.\n\n60 ml * (1.0 liter/1000 ml) * (0.250 mole/liter HCl) = .015 mole HCl\n\n60 ml * (1.0 liter/1000 ml) * (0.125 mole/liter NaOH) = .0075 moles NaOH\n\nThat means you will have 0.015 - 0.075 = 0.075 moles of HCl left over after the neutralization.\n\nI leave it for you to calculate the pH of a solution of 0.075 moles of HCl in 120 ml of solution.  Remember that pH = -log(base 10)[H+]
911223
Take 1ft. as your height of the capillary tube - when placed in a trough of water.\nAdd eac detergent each time (separately), notice the rise in level - looong time consuming, since u need to change water for each detergent, but a short way is:\nFrom ur experience, see which is the strongest detergent, in order of strength - separate them by 0.5 gradation - meaning, if ur tube gradation is 1mm, the diff. in level between the consecutive order (of detergents) will be 0.5 mm respectively.
857742
Any acid.  Vinegar is the best probably.
198886
somethings wrong here dude. you cant get something more concentrated. like just think about it, the only way you could make it more concentrated is to take otu water from it, which isnt what youre supposed to do here.\ni think maybe you got the question backwards. from 100mL of 0.92M NaCl, make 0.39M solution. it doesnt matter, its same thing. which is \n0.92 x 0.1 = 0.39 x v2      \nv2 = 236mL. so basically you take your 100mL and add 136mL of water to it  so the new volume is 236mL. and the new concentration would be 0.39M
993248
HCI hydrogen chloride \nto make HCI u have to mix Sulfuric Acid and Sodium
607804
The easiest way to prepare an insoluble salt is to produce it from two aqueous solutions each containing an ion present in the salt!\n\nSo, you'll just mix two solutions, and the insoluble salt appears! Of course, if you intend to prepare a certain amount/ mass of the salt, you may need to carry out some calculations to determine the necessary concentrations/ volumes of the 2 initial solutions. \n\nYou'll also get another product in the reaction, but it should be soluble and will not contaminate the desired product (the insoluble salt).  \n\nYou may refer to this website to gain a better understanding...\nhttp://www.wpbschoolhouse.btinternet.co.uk/page01/AqueousChem/AqueousChem.htm#salts [Refer to point (4)]
493490
Buffer capacity is defined as the moles of acid or base necessary to change the pH of one liter of solution by one unit.\n\nBuffer Capacity = (number of moles of OH- or H3O+ added)     divided by ( pH change) x (volume of buffer in L)\n\n-Buffering capacity increases as the molar concentration (molarity)of the buffer salt/acid solution increases.\nThat's only partially true.The pH of a buffer solution is independent of buffer concentration only if there is a sufficient quantity present. A buffer consists of a solution of a weak acid (base) and the salt of weak acid (base). The addition of either acid or base converts acid(base) to the acid (base) salt, or the salt back to acid (base) when another acid or base is added to the solution.If there is not enough buffer present compared to the added acid or base, the buffer can be swamped, and the pH will change significantly. So it is only partially true that the buffer is independent of the buffer concentration.If the amount of acid or base added, is small compared to the buffer concentration then the pH is not sensitive to the amount of buffer present\n\n-the closer the bufferd pH is to the pKa,the greater the buffering capacity\n\nThe solution is at a maximum buffer capacity (the minimum dpH/dn) when pH = pKa, as can be seen algebraically: \nKa = [H3O+][A-]/[HA]; [H3O+] = Ka[HA]/[A-] \n\nlog [H3O+] = log Ka + log ([HA]/[A-]) \n\npH = pKa - log([HA]/[A-]) \n\nThis rearranged form of the ionization constant of an acid is sometimes called the Henderson-Hasselbalch equation. \n\nSince pH = pKa if and only if [HA] = [A-], maximum buffer capacity of a conjugate acid-base pair is always found at pH = pKa. Reasonable buffer capacity is still obtained if the ratio of [HA] to [A-] is somewhere between 0.1 and 10, but outside this range it decreases rapidly. A chemical buffer can be effective about pH = pKa +/- 1. To make up a buffer solution one would first choose a reasonable acid-base conjugate pair on the basis of its pKa value and any other chemical constraints (one would not use the HCN/CN- pair, for example, in biological systems because cyanide is toxic). One would then adjust the concentrations such that both are reasonable (say 10-3 molar or above) while the desired pH is still obtained. The conjugate acid-base pairs glycinium ion/glycine, hydrogen citrate/citrate, dihydrogen phosphate/monohydrogen phosphate, and boric acid/hydrogen borate are often used for the preparation of aqueous buffers.
885235
A thorough perusal of your textbook should help with your problem.  Oh, wait--sorry, I forgot--you aren't interested in LEARNING how to calculate the pH of your sodium acetate solution, just in getting the quick fix answer from someone who has.
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.
665494
It can be made by dehydration of acrylamide\nCH2=CH-C(O)NH2 --> CH2=CH-CN + H2O\nP2O5 is a good reagent for that.
311730
I think you should keep it simple yet smart. First get a piece of pure elemental Sodium and then let it drop on warm water. Afterwards explain how the chemical reaction released Sodium Oxide and Hydrogen Gas.
355477
Buffers resist change in pH. This is generally a result of the buffer containing a mix of a weak acid and the salt of the conjugate base. \n\nWhen you add NaOH to the buffer, it dissociates into Na+ and OH-. The OH- reacts with the H+ ions which the weak acid has dissociated into (along with the conjugate base) up to the acid dissociation constant, Ka, of the acid.\n\nThis shifts the equilibrium of the dissociation reaction of the acid, and more dissociates, maintaining a relatively stable pH, as compared to when the NaOH is added to the distilled water. (If you had instead added a base, it would instead have shifted the equilibrium of the base, and, instead, more of the base would have reacted with the free H+, again, keeping the pH relatively constant)