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Buffers with Specific Ionic Strength 1. How many mL of 12.0 M acetic acid and how many grams of sodium acetate (FW = 82 g/mol) are needed to prepare a.

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Presentation on theme: "Buffers with Specific Ionic Strength 1. How many mL of 12.0 M acetic acid and how many grams of sodium acetate (FW = 82 g/mol) are needed to prepare a."— Presentation transcript:

1 Buffers with Specific Ionic Strength 1

2 How many mL of 12.0 M acetic acid and how many grams of sodium acetate (FW = 82 g/mol) are needed to prepare a 500 mL buffer at pH 5.0, and having an ionic strength of 0.2. k a = 1.8*10 -5 We need to find the concentration of the salt:  = ½  C i Z i = ½ (C Na + * C OAc - * 1 2 ) C Na + = C OAc = ½ (2C Na + ) C Na + = C OAc - = 0.2 M = C NaOAc mmol NaOAC = 0.2*500 = 100 mg NaOAc = 100*82 = 8200 mg or 8.2 g 2

3 HOAc  H + + OAc - We can now find the concentration of the acid where: 1.8*10 -5 = *0.2/[HOAc] [HOAc] = 0.2/1.8 = 0.11 M mmol HOAc = 0.11*500 = *V HOAc = 55.6 V HOAc = 4.6 mL 3

4 How many grams of Na 2 CO 3 (FW = 106 g/mol) and how many grams of NaHCO 3 (FW = 84 g/mol) are needed to prepare a 1000 mL buffer at pH 10.0, and having an ionic strength of 0.2. k a2 = 4.8* HCO 3 -  CO H + 4.8* = [CO 3 2- ]/[HCO 3 - ] [HCO 3 - ] = 2.1[CO 3 2- ]  = ½  C i Z i 2 4

5 In NaHCO 3, C Na + = C HCO3 - In Na 2 CO 3, C Na + = 2C CO = ½ (C Na + *1 2 + C HCO3 - *1 2 + C Na + *1 2 + C CO3 2- *2 2 ) 0.2 = ½ (C HCO3 - *1 2 + C HCO3 - * C CO3 2- *1 2 + C CO3 2- *2 2 ) However, [HCO 3 - ] = 2.1[CO 3 2- ] 0.2 = ½ (2* 2.1[CO 3 2- ] + 2 C CO3 2- *1 2 + C CO3 2- *2 2 ) [CO 3 2- ] = M [HCO 3 - ] = 2.1[CO 3 2- ] = 2.1* = M g Na 2 CO 3 = *1000 * 106= 4.16 g NaHCO 3 = *1000*84 = 6.92g 5

6 How many mL of 12.0 M acetic acid and how many grams of sodium acetate (FW = 82 g/mol) and how many grams of NaNO 3 (FW = 85 g/mol) are needed to prepare a 500 mL buffer at pH 5.0, and a salt concentration of 0.1 M and an ionic strength of k a = 1.8*10 -5 We need to find the concentration of the salt:  = ½  C i Z i 2  = {(  NaNO3 )+(  NaOAc )} 0.15 = ½ {(C Na + * C NO3 - * 1 2 )+ (0.1* *1 2 )} C Na + = C NO3 - = C NaNO = ½ {(2C NaNO3 ) + 0.2} C NaNO3 = 0.05 M g NaNO 3 = 0.05*500*85 = 2125mg = g 6

7 C NaOAc = 0.1 M mmol NaOAC = 0.1*500 = 50 mg NaOAc = 500*82 = 4100 mg or 4.1 g HOAc  H + + OAc - We can now find the concentration of the acid where: 1.8*10 -5 = *0.1/[HOAc] [HOAc] = 0.1/1.8 = M mmol HOAc = 0.056*500 = *V HOAc = 27.8 V HOAc = 5.6 mL 7


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