Determine the mass of NH2CH2CO2Na and the volume of 0.150 M HCl solution required to make 1.35 L of a buffer solution at pH=10.75 and at a total concentration of 0.33 M. pKa(NH2CH2CO2H)=9.88.
a.Mass of NH2CH2CO2Na:
b. Volume of HCl solution:
Hi Im sooo confused.
9.88=10.75+log(0.15/HCL)
-0.87=log(0.15/HCL)
HCL=0.135M
Mass=0.135M(1.35L)(36.46g/mol)
Mass=6.64g
but its wrong what???why?
To determine the mass of NH2CH2CO2Na and the volume of HCl solution required, we need to use the Henderson-Hasselbalch equation to calculate the required concentration of the acid and its corresponding volume.
First, let's find the concentration of the conjugate acid NH2CH2CO2H. We know that pH = pKa + log(base/acid). In this case, the base (NH2CH2CO2-) is the salt NH2CH2CO2Na and the acid is NH2CH2CO2H.
pH = 10.75
pKa = 9.88
10.75 = 9.88 + log(Na+/NH2CH2CO2H)
Rearranging the equation:
log(Na+/NH2CH2CO2H) = 10.75 - 9.88
log(Na+/NH2CH2CO2H) = 0.87
Now, we can find the ratio of Na+/NH2CH2CO2H using the antilog function:
Na+/NH2CH2CO2H = 10^(0.87)
Na+/NH2CH2CO2H = 6.76
Since the total concentration of the buffer solution is given as 0.33 M, we can set up the following equation using the law of mass action:
[Na+] + [NH2CH2CO2H] = 0.33 M
Substituting the ratio we found:
6.76[NH2CH2CO2H] + [NH2CH2CO2H] = 0.33 M
7.76[NH2CH2CO2H] = 0.33 M
[NH2CH2CO2H] = 0.33 M / 7.76
[NH2CH2CO2H] = 0.0425 M
Now that we have the desired concentration of NH2CH2CO2H, we can determine the volume of HCl solution needed to create the buffer.
[HCl] = 0.15 M
Using the formula [volume] = [moles] / [concentration], we can calculate the moles of NH2CH2CO2H and the corresponding volume of HCl solution:
moles of NH2CH2CO2H = [NH2CH2CO2H] * [volume]
moles of NH2CH2CO2H = 0.0425 M * 1.35 L
moles of NH2CH2CO2H = 0.057375 mol
moles of HCl = moles of NH2CH2CO2H
Using the equation [moles] = [concentration] * [volume]:
0.15 M * volume = 0.057375 mol
volume = 0.057375 mol / 0.15 M
volume = 0.3825 L
Therefore, the volume of HCl solution required to make 1.35 L of the buffer solution is 0.3825 L.
Now, let's calculate the mass of NH2CH2CO2Na using the molar mass of NH2CH2CO2Na (84.06 g/mol) and the calculated concentration of Na+:
mass of NH2CH2CO2Na = [Na+] * [volume] * [molar mass]
mass of NH2CH2CO2Na = 6.76 * 1.35 L * 84.06 g/mol
mass of NH2CH2CO2Na = 766.71 g
Therefore, the mass of NH2CH2CO2Na required to make 1.35 L of the buffer solution is 766.71 g.
It appears that the calculation you provided is incorrect. Please double-check your equations and calculations to ensure accuracy.