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# 1- Say- for example- that you had prepared a Buffer C- in which you mi.docx

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# 1- Say- for example- that you had prepared a Buffer C- in which you mi.docx

1.      Say, for example, that you had prepared a Buffer C, in which you mixed 8.203 g of sodium acetate, NaC 2 H 3 O 2 , with 100.0 mL of 1.0 M acetic acid.
a.       What would be the initial pH of Buffer C?
?- i got pH 4.74
b.       If you add 5.0 mL of 0.5 M NaOH solution to 20.0 mL each of Buffer B and Buffer C, which buffer’s pH would change less? Calculate the pH after the addition of the base to both buffers B and C. (buffer B was made with 0.735g sodium Acetate with 50ml of 10M acetic Acid)
c.    If you wanted to carry out an experiment at ‘physiological pH’ (7.4) could you suggest an appropriate buffer system? (Hint: The blood buffer system is incredibly complex, and does not make for a good buffer system in the lab)
Solution
pH of buffer = pKa + log[salt]/[Acid]
pH = 4.76+log((8.203/82.03*1000/100)/(100*1/1000))
pH = 5.76
pH of buffer B = pKa + log[salt]/[Acid]
pH = 4.76 +log((0.735/82.03*1000/100)/(20*10/1000))
pH = 4.411
if we add NaOH to the above buffer solutions then,
pH of Buffer C = pKa + log[Salt + base]/[Acid - Base]
pH = 4.76+log(((8.203/82.03*1000/100)+(5*0.5/1000))/((100*1/1000)-(5*0.5/1000)))
pH = 5.77
pH of Buffer B = pKa + log[Salt + base]/[Acid - Base]
pH = 4.76 +log(((0.735/82.03*1000/100)+(0.5*5/1000))/((20*10/1000)-(0.5*5/1000)))
pH = 4.428
pH Change for buffer C = 5.77 - 5.76 = 0.06
pH change for buffer B = 4.428 - 4.411 = 0.017
Ph change is higher for buffer B
.

1.      Say, for example, that you had prepared a Buffer C, in which you mixed 8.203 g of sodium acetate, NaC 2 H 3 O 2 , with 100.0 mL of 1.0 M acetic acid.
a.       What would be the initial pH of Buffer C?
?- i got pH 4.74
b.       If you add 5.0 mL of 0.5 M NaOH solution to 20.0 mL each of Buffer B and Buffer C, which buffer’s pH would change less? Calculate the pH after the addition of the base to both buffers B and C. (buffer B was made with 0.735g sodium Acetate with 50ml of 10M acetic Acid)
c.    If you wanted to carry out an experiment at ‘physiological pH’ (7.4) could you suggest an appropriate buffer system? (Hint: The blood buffer system is incredibly complex, and does not make for a good buffer system in the lab)
Solution
pH of buffer = pKa + log[salt]/[Acid]
pH = 4.76+log((8.203/82.03*1000/100)/(100*1/1000))
pH = 5.76
pH of buffer B = pKa + log[salt]/[Acid]
pH = 4.76 +log((0.735/82.03*1000/100)/(20*10/1000))
pH = 4.411
if we add NaOH to the above buffer solutions then,
pH of Buffer C = pKa + log[Salt + base]/[Acid - Base]
pH = 4.76+log(((8.203/82.03*1000/100)+(5*0.5/1000))/((100*1/1000)-(5*0.5/1000)))
pH = 5.77
pH of Buffer B = pKa + log[Salt + base]/[Acid - Base]
pH = 4.76 +log(((0.735/82.03*1000/100)+(0.5*5/1000))/((20*10/1000)-(0.5*5/1000)))
pH = 4.428
pH Change for buffer C = 5.77 - 5.76 = 0.06
pH change for buffer B = 4.428 - 4.411 = 0.017
Ph change is higher for buffer B
.

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### 1- Say- for example- that you had prepared a Buffer C- in which you mi.docx

1. 1. 1. Say, for example, that you had prepared a Buffer C, in which you mixed 8.203 g of sodium acetate, NaC 2 H 3 O 2 , with 100.0 mL of 1.0 M acetic acid. a. What would be the initial pH of Buffer C? ?- i got pH 4.74 b. If you add 5.0 mL of 0.5 M NaOH solution to 20.0 mL each of Buffer B and Buffer C, which buffer’s pH would change less? Calculate the pH after the addition of the base to both buffers B and C. (buffer B was made with 0.735g sodium Acetate with 50ml of 10M acetic Acid) c. If you wanted to carry out an experiment at ‘physiological pH’ (7.4) could you suggest an appropriate buffer system? (Hint: The blood buffer system is incredibly complex, and does not make for a good buffer system in the lab) Solution pH of buffer = pKa + log[salt]/[Acid] pH = 4.76+log((8.203/82.03*1000/100)/(100*1/1000)) pH = 5.76 pH of buffer B = pKa + log[salt]/[Acid] pH = 4.76 +log((0.735/82.03*1000/100)/(20*10/1000)) pH = 4.411 if we add NaOH to the above buffer solutions then, pH of Buffer C = pKa + log[Salt + base]/[Acid - Base] pH = 4.76+log(((8.203/82.03*1000/100)+(5*0.5/1000))/((100*1/1000)-(5*0.5/1000))) pH = 5.77
2. 2. pH of Buffer B = pKa + log[Salt + base]/[Acid - Base] pH = 4.76 +log(((0.735/82.03*1000/100)+(0.5*5/1000))/((20*10/1000)-(0.5*5/1000))) pH = 4.428 pH Change for buffer C = 5.77 - 5.76 = 0.06 pH change for buffer B = 4.428 - 4.411 = 0.017 Ph change is higher for buffer B