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By dilution of 1 part 0.50M buffer with 49 parts of water, giving 50 parts of the desired 0.010M = 10 mM Phosphate buffer
Cfjn
calculat the [H3o+] 7.5 =-log 10 h+
Take 333 milliliters of your stock solution and dilute it to 1L with water.
For pH = 7,0: 756 mL disodium hydrogen phosphate (Na2HPO4) solution 0,1 M (14,2 g/L) + 244 mL hydrochloric acid (HCl) solution 0,1 M
By dilution of 1 part 0.50M buffer with 49 parts of water, giving 50 parts of the desired 0.010M = 10 mM Phosphate buffer
Cfjn
calculat the [H3o+] 7.5 =-log 10 h+
Take 333 milliliters of your stock solution and dilute it to 1L with water.
For pH = 7,0: 756 mL disodium hydrogen phosphate (Na2HPO4) solution 0,1 M (14,2 g/L) + 244 mL hydrochloric acid (HCl) solution 0,1 M
See the Web Links to the left of this answer.I especially like the Smith.edu link -- it has complete and very useful description of how to prepare a buffer.Use the Henderson-Hasselbach equation:pH = pKa + log [A-]/[HA]where HA is the protonated form of the weak acid, A- is the salt (dissociated acid, or in other words, its conjugate base), and the pKa is the strength of the acid.What this says is that the pH that you want your buffer to be depends on two things:-- the pKa of the weak acid you are using (see reference tables under the Web Links to the left)-- and the RATIO of the concentration of the acid and the salt that you add to the solution.The pH of the buffer does not depend on the actual concentration of the buffer, but on the ratio of the two parts.The buffer capacity depends on two things -- how close to the pKa the pH of the buffer actually is (it should be within 1-2 pH units), and what the total concentration of the buffer is.For instance if you have 0.001 M acetic acid and 0.001 M sodium acetate, the resulting buffer will have the exact same pH as a buffer made with 0.1 M acetic acid and 0.1 M sodium acetate (because the ratio is 1 to 1, the pH = pKa = 4.76). However, the 0.1 M buffer will have a much larger buffer capacity, and will much better resist changes in pH upon the addition of a strong acid or base.
Decide on the concentration of the buffer, use 1L to be simple PH for your buffer should be within one pH unit from the pKa of the acid/conjugate base use Henderson Hasselbalch Equation pH = pKa + log ([Base]/[Acid]) For a 1 M buffer [Acid] + [Base] = 1
For the preparation of a solution with the pH=7,00:Add 29,1 mL sodium hydroxide solution 0,1 M to 50 mL potassium dihydrogen phosphate solution 0,1 M.
1 M Sodium Phosphate Buffer Stock Solution (1 liter) Protocol # Solution A: Dissolve 138.0 g NaH2PO4?H2O in 1 liter dH2O (pH 7.0). # Solution B: Dissolve 142.0 g Na2HPO4 in 1 liter dH2O (pH 7.0). # Mix 423 ml Solution A with 577 ml Solution B. # Autoclave and store at room temperature.
Phosphate buffer pH 6.8 preparation protocol below Stock solutions: 0.2M dibasic sodium phosphate (1 liter) Na2HPO4*12H2O (MW=358.14) --------71.64gm + dH2O to make 1 liter (Solution X) 0.2M monobasic sodium phosphate (1 liter) NaH2PO4*H2O (MW=138.01) --------27.6gm + dH2O to make 1liter (Solution Y) Working buffer: 0.1M (1 liter) pH 6.8 245 ml solution X + 255 ml solution Y ( filled up to 1 liter with dH2O)
To prepare the buffer using solid form reagents, prepare a 0.1 M ammonium acetate solution by dissolving 7.7 g ammonium acetate in a 1000 ml water. Adjust 1 L of this solution to pH 4.5 by adding acetic acid (about 8 ml) and 5 ml of 1 M p-TSA (equivalent to 5 mM p-TSA).
Britton-Robinson buffer is a "universal" pH buffer used for the range pH 2 to pH 12. Universal buffers consist of mixtures of acids of diminishing strength (increasing pKa) so that the change in pH is approximately proportional to the amount of alkali added. It consists of a mixture of 0.04 M H3BO3, 0.04 M H3PO4 and 0.04 M CH3COOH that has been titrated to the desired pH with 0.2 M NaOH. Britten and Robinson also proposed a second formulation that gave an essentially linear pH response to added alkali from pH 2.5 to pH 9.2 (and buffers to pH 12). This mixture consists of 0.0286 M citric acid, 0.0286 M KH2PO4, 0.0286 M H3BO3, 0.0286 M veronal and 0.0286 M HCl titrated with 0.2 M NaOH.