 # Hcl note you may ignore the slight change in volume

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HCl? Note: You may ignore the slight changein volume caused by the HCl addition.nofNa HPOmol240072 01 00072=´ =()...(2.32)nofNaH POmol240028 01 00028=´ =()...(2.33)nM vofHCl molmol()= ´ = ´=()2 00010002..(2.34)nofNa HPO afterHCladdition mol2400072 0002 00052()-=:...(2.35)nofNaH PO afterHCladdition mol2400028 0002 00048()+=:...(2.36)pH of buffer after HCl addition and conversionof n intoMthrough Eq. 2.1:pH=+[][]=67100520048674.log...(2.37)Note: The ratio of Ato AH stays the samewhether concentrations or amounts are inserted,since units are canceled, so technically n does notneed to be converted intoMto obtain the correctresult.Example C3Prepare 250 mL of 0.1Macetatebuffer with pH 5. The pKaof acetic acid is 4.76(see Table2.2). The molecular weights of aceticacid and sodium acetate are 60.06 and 82.03 g/mol, respectively.SolutionThis example matches the tasks athand in a lab better than ExampleC1. One needsa buffer to work at a certain pH, looks up thepKavalue, and decides on the molarity and vol-ume needed. The molarity of a buffer equals[A] + [AH]. To solve ExampleC3, one of thoseconcentrations needs to be expressed in terms ofthe other, so that the equation only contains oneunknown quantity. For this example, it will be[A], but the results would be the same if [AH]had been chosen. Together with the target pH(5) and the pKa(4.76), the values are insertedinto Eq. 2.25:MolarityofbuffermolLAHAæèçöø÷=[]+-[ ](2.38)01. [ ]=+[]AAH-(2.39)[AAH]. –=[]01(2.40)C. Tyl and B.P. Ismail
29There are three ways to prepare such a buffer:1. Prepare 0.1Macetic acid and 0.1Msodium ace-tate solutions. For our example, 1 L will be pre-pared. Sodium acetate is a solid and can beweighed out directly using Eq. 2.5. For aceticacid, it is easier to pipet the necessary amount.Rearrange Eq. 2.8to calculate the volume ofconcentrated acetic acid (density = 1.05) neededto prepare a volume of 1 L. Then express themass through Eq. 2.5.mM vofsodiumacetate gMW( )= ´ ´(2.5)mofsodiumacetate gmolLLgmolg( )=æèçöø÷´( )´æèçöø÷=( )0118282..(2.50)vofaceticacid mLm gdgmL()=( )æèçöø÷(2.51)vMvofaceticacid mLmolLLMWgmoldgmL()=æèçöø÷´( )´æèçöø÷æèçöø÷(2.52)vofaceticacid mLgmolmolLLgm()=æèçöø÷´æèçöø÷´( )6006011105...LmLæèçöø÷=()572.(2.53)Dissolving each of these amounts of sodiumacetate and acetic acid in 1 L of water gives two1  L stock solutions with a concentration of0.1M. To calculate how to mix the stock solu-tions, use their concentrations in the buffer, i.e.,0.0365 as obtained from Eqs. 2.48and2.49  –0.0365molLæèçöø÷for acetic acid and 0.0635molLæèçöø÷for sodium acetate and Eq. 2.15:vMofaceticacidstocksolution LofaceticacidinbuffermolL( )=æèçöø÷´( )æèçöø÷vMofbuffer LofstocksolutionmolL(2.54)vofaceticacidstocksolution LL( )=´=( )00365 025010091....(2.55)vofsodiumacetatestocksolution LL( )=´=( )00635 025010159....(2.56)Combining the 0.091 L of acetic acid stock and0.159  L of sodium acetate stock solution gives0.25 L of buffer with the correct pH and molarity.2. Directly dissolve appropriate amounts of bothcomponents in the same container. Equations 2.48and2.49yield the molarities of acetic acid andsodium acetate in a buffer, i.e., the moles per 1

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Term
Summer
Professor
Lam An Vu
Tags
Accuracy and precision, Laboratory glassware, Pipette, Mechanical Pipettes, Laboratory Standard Operating Procedures
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