What volume of 1.0 M NaOH must be added to 1.5 L of 1.20 M CH3COOH (ka=1.8x10^-5) to make the solution a buffer for pH 4.50?
I do not understand how to determine the mols of NaOH. I tried to use the equation where:
[H3O+] = ka (nHA/nA-)
but it is not giving me the correct answer of 0.657 L
How many mols CH3COOH (HAc) do you have? That's M x L = 1.2 x 1.5 = 1.8
.........HAc + OH^- ==> Ac^- + H2O
I........1.8...0........0.......0
add............x..................
C.........-x..-x........+x......
E.......1.8-x...0........x
Now substitute the E line into the Henderson-Hasselbalch equation and solve for x = OH^ = mols OH to be added. I get 0.657 mols.
Then M NaOH = mols NaOH/L NaOH.
or L NaOH = mols NaOH/M NaOH
L NaOH = 0.657 mols/1.0 M = 0.657 L. :-)
How do you get 0.657 mols? I solved for x and got 0.6529 mols
the preciseness in the original question is two sig digits...
however, I suspect your error is in the antilog of pH look there.
Here is my work and I keep getting the hundredth value to be incorrect:
10^-4.5 = [1.8E-5 (1.8-x)]/x
10^-4.5 = [3.24E-5 - 1.8E-5]/x
10^-4.5x = 3.24E-5 - 1.8E-5x
4.96E-5x = 3.24E-5
x= 0.653
10^-4.5 = [1.8E-5 (1.8-x)]/x
10^-4.5 = [3.24E-5 - 1.8E-5x]/x
x(10^-4.5 +1.8E-5)=3.24E-5
x(10^.5+1.8)=3.24
x*4.96=3.24
x=.655
To determine the volume of 1.0 M NaOH required to make the solution a buffer for pH 4.50, you need to consider the Henderson-Hasselbalch equation for a buffer solution:
pH = pKa + log([A-]/[HA])
Where [A-] is the concentration of the conjugate base (acetate ion, in this case) and [HA] is the concentration of the acid (acetic acid, CH3COOH).
Let's break down the problem step by step:
1. Determine the pKa value:
Given that the Ka value is 1.8x10^-5, you can calculate the pKa value using the equation pKa = -log(Ka). In this case, pKa is equal to 4.74.
2. Determine the concentration of the conjugate base ([A-]):
The buffer solution should have a pH of 4.50. Using the Henderson-Hasselbalch equation, you can rearrange it to find the ratio [A-]/[HA]:
[A-]/[HA] = antilog (pH - pKa)
Here, antilog refers to raising 10 to the power of the value inside the parentheses. Therefore, [A-]/[HA] = antilog (4.50 - 4.74).
Calculating antilog (4.50 - 4.74), you'll find that [A-]/[HA] = 0.427.
3. Determine the concentration of the acid ([HA]):
Given that the initial concentration of acetic acid is 1.20 M and the total volume is 1.5 L, you can calculate the moles of acetic acid present in the solution:
moles of CH3COOH = concentration × volume = 1.20 M × 1.5 L = 1.80 moles
4. Determine the moles of the conjugate base ([A-]):
Using the ratio [A-]/[HA] = 0.427, you can determine the moles of the conjugate base:
moles of A- = 0.427 × 1.80 moles = 0.7716 moles
5. Determine the amount of NaOH required:
Since 1 mole of NaOH reacts with 1 mole of acetic acid (CH3COOH) to form 1 mole of sodium acetate (CH3COONa) in the balanced equation:
CH3COOH + NaOH → CH3COONa + H2O
The moles of NaOH required will be equal to the moles of acetic acid:
moles of NaOH = moles of CH3COOH = 0.7716 moles
6. Finally, determine the volume of 1.0 M NaOH required:
To calculate the volume, you need to use the definition of molarity: Molarity = moles/volume.
Therefore, volume = moles/Molarity = 0.7716 moles / 1.0 M = 0.7716 L (or 0.657 L when rounded to three significant figures).
So, the correct answer for the volume of 1.0 M NaOH required to make the solution a buffer for pH 4.50 is approximately 0.657 L.