What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250?
I assume you mean M and not m. m stands for molality. M is for molarity. I will use HAc for
millimols HAc = mL x M = 200 x 0.425 = 85
...............HAc + NaOH ==> NaAc + H2O
I...............85.........0................0............0
add ......................x.....................................
C.............-x.........-x.................x.............x
E............85-x........x.................x.............x
Plug the E line into the Henderson-Hasselbalch equation and solve for x =- millimols NaOH to be added. Then M = millimols/mL. You know M and millimols, solve for mL.
To solve this question, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
First, let's find the concentration of acetic acid (HA) and its conjugate base (A-) in the buffer solution.
Given:
- Volume of acetic acid (HA) = 200.0 mL = 0.200 L
- Concentration of acetic acid (HA) = 0.425 M
The moles of acetic acid (HA) can be calculated as follows:
moles of HA = concentration × volume = 0.425 M × 0.200 L = 0.085 moles
Now, let's consider the Henderson-Hasselbalch equation rearranged to solve for the ratio [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
Since we want a pH of 4.250, and the pKa of acetic acid (CH3COOH) is given as 1.75 x 10^-5:
[A-]/[HA] = 10^(4.250 - (-log10(1.75 x 10^-5))) = 10^(4.250 + 4.756) = 10^9.006
Now, let's solve for the concentration of the conjugate base A-. We know that the moles of A- and HA are in a 1:1 ratio (since they are produced by the same number of moles of NaOH), so the concentration of A- in the final solution will be the same as that of HA.
[A-] = 0.085 moles / 0.200 L = 0.425 M
Now we need to find the concentration and volume of the NaOH solution we need to add to the acetic acid solution to produce the desired buffer.
Let's assume that the volume of NaOH added is V mL, and the molarity of NaOH is C M.
The moles of NaOH added can be calculated as follows:
moles of NaOH = concentration × volume = C M × V mL = C × V moles
Since the moles of NaOH and A- are in a 1:1 ratio (according to the balanced equation), the concentration of A- can now be expressed as:
[A-] = C × V moles / (0.200 L + V mL)
Given the desired concentration of A- (0.425 M), we can set up the following equation:
0.425 M = C × V moles / (0.200 L + V mL)
Now, we need to find the pH at which A- and HA are in a 1:1 ratio. This can be determined by substituting [A-] = [HA] into the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Substituting the values:
4.250 = -log10(1.75 x 10^-5) + log(1)
Since log(1) = 0, we can simplify to:
4.250 = -log10(1.75 x 10^-5)
Taking the antilog of both sides:
10^4.250 = 1.75 x 10^-5
Now, let's solve for C × V:
C × V = (0.200 L + V mL) × 1.75 x 10^-5
C × V = 3.5 x 10^-6 L + 1.75 x 10^-5 V mL
Now, substitute the value of C × V into the equation:
0.425 M = (3.5 x 10^-6 L + 1.75 x 10^-5 V mL) / (0.200 L + V mL)
Cross multiplying:
0.425 M × (0.200 L + V mL) = 3.5 x 10^-6 L + 1.75 x 10^-5 V mL
0.085 L + 0.425 V mL = 3.5 x 10^-6 L + 1.75 x 10^-5 V mL
0.425 V mL - 1.75 x 10^-5 V mL = 3.5 x 10^-6 L - 0.085 L
0.425 V mL - 1.75 x 10^-5 V mL = - 0.085 L + 3.5 x 10^-6 L
0.425 V mL - 1.75 x 10^-5 V mL = - 0.085 L + 3.5 x 10^-6 L
0.425 V mL - 1.75 x 10^-5 V mL = - 0.085 L + 3.5 x 10^-6 L
Combining like terms:
0.40725 V mL = -0.0815 L
Dividing both sides by 0.40725:
V mL = -0.0815 L / 0.40725
V mL = -0.2 L
Since it is not physically possible to have negative volume, we can conclude that it is not possible to create a buffer of pH = 4.250 with the given concentrations.