You need to be a little bit more specific
Answer:
All I want is to fly with you
All I want is to fall with you
So just give me all of you
It feels impossible
It's not impossible
Is it impossible?
Say that it's possibleExplanation:
Explanation:
huffy Cranbrook cruiser cñhc lol oh oh it o hj p
<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
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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>
