Explanation:
It’d be a good question to pose to Norman Rockwell, that famed painter of quaint, funny scenes depicting mid-20th-century American life. His works were reproduced ceaselessly on magazine covers in the 1930s, ’40s, and ’50s—and their appeal was immense. By the 1940s, Time magazine had already christened Rockwell as “probably the best-loved U.S. artist alive,” while the New York Times had affectionately compared his paintings to Mark Twain’s novels.
<span>pH = pKa + log ([R-]/[RH])
Where pH is the pH of the buffer, [R-] is the concentration of the basic species, and [RH] is the concentration of the acidic species.
At pH 2.4, the amino group on glycine (pKa = 9.6) will be, for accounts and purposes, 100% protonated. This means our buffer will be dealing with the two ionic forms of the carboxyl group (pKa = 2.4).
When pH = pKa, the two species are in equilibrium. This can be seen using the HH equation:
2.4 = 2.4 + log ([R-]/[RH])
0 = log ([R-]/[RH])
1 = ([R-]/[RH])
[RH] = [R-]
Now we add in another equation, our conservation of mass.
M = [RH] + [R-]
where M is the molarity of the buffer
But since [RH] = [R-]:
M = 2 [RH]
0.2 = 2 [RH]
And we wind up with:
[RH] = [R-] = 0.1 M
Now to figure out the moles of each needed, we multiply by the volume of the buffer.
0.1 M * 0.1 L = 0.01 mol
This shows that to make 100 ml of 0.2 M glycine buffer, we'll need 0.01 mol of each species.
0.01 mol of 0.5 M HCl:
0.5 mol HCl / 1 L = 0.01 mol / v
solve for v
v/1 = 0.01 / 0.5 ==> v = 0.02 L or 20 mL
weight of glycine:
MW: 75.07 g/mol
0.01 mol glycine * (75.07g glycine / 1 mol) = 0.75 g glycine
And there's your answer
--------
To make this buffer you would add 0.75g glycine to 20 mL of 0.5 M HCl and fill with water until a 100mL volume was achieved.</span>
Answer:
Jus try ur best and train harder for the next one
Explanation:
