I got pH = 5.3
A 1.00 L buffer solution is 0.250 M in HC2H3O2 and 0.250 M in NaC2H3O2. Calculate the pH of the solution after the addition of 0.150 mol of solid NaOH. Assume no volume change upon the Additon o f NaOH. The Ka value for HC2H3O2 at 25 C is 1.8 x 10^-5.
Chemistry-Dr.Bob222 - DrBob222, Friday, April 29, 2016 at 2:38pm
HC2H3O2 = HAc
NaC2H3O2 = NaAc
millimols HAc = mL x M = approx 250
mmols NaAc = 250
0.150 mol NaOH = 150 mmol
......HAc + OH^- ==> Ac^- + H2O
I.....250....0.......250.......
add........150.................
C...-150..-150.......+150
E.....100...0........400
Plug the E line into the HH equation and solve for pH.
That's ok but I believe you are allowed another significant figure. That 5 comes from the log. I obtained 5.34.
To solve for the pH of the solution after the addition of NaOH, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentration of the acid and its conjugate base.
First, we need to calculate the new concentration of the acetic acid (HC2H3O2) and its conjugate base (C2H3O2^-) after the addition of NaOH.
Initially, the buffer solution is 0.250 M in both HC2H3O2 and NaC2H3O2. Since there is no volume change upon the addition of NaOH, the final volume of the solution is still 1.00 L.
The initial concentration of HC2H3O2 is 0.250 M. Since the volume is 1.00 L, the initial number of moles of HC2H3O2 is:
0.250 M x 1.00 L = 0.250 mol
The initial concentration of NaC2H3O2 is also 0.250 M. Therefore, the initial number of moles of C2H3O2^- is:
0.250 M x 1.00 L = 0.250 mol
The addition of 0.150 mol of NaOH converts 0.150 mol of HC2H3O2 into C2H3O2^- and water. Therefore, the final number of moles of C2H3O2^- is:
0.250 mol + 0.150 mol = 0.400 mol
Since the volume is still 1.00 L, the final concentration of C2H3O2^- is:
0.400 mol / 1.00 L = 0.400 M
Now, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([C2H3O2^-] / [HC2H3O2])
The pKa value for HC2H3O2 is given as 1.8 x 10^-5.
Substituting the values into the equation, we have:
pH = -log(1.8 x 10^-5) + log(0.400 M / 0.250 M)
Simplifying the equation:
pH = -log(1.8 x 10^-5) + log(1.6)
Using the logarithmic properties, the equation becomes:
pH = -(-log(1.8 x 10^-5) - log(1.6))
Calculating the values inside the parentheses and simplifying, we get:
pH = -(-4.74 - 0.204)
pH = -(-4.94)
pH = 4.94
Therefore, the pH of the solution after the addition of 0.150 mol of solid NaOH is approximately 4.94.
To calculate the pH of the solution after the addition of solid NaOH, you need to use the Henderson-Hasselbalch equation. This equation is:
pH = pKa + log([A-]/[HA])
Where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
First, let's calculate the concentrations of the acid (HAc) and the conjugate base (Ac-) after the addition of NaOH.
Initially, the solution is 0.250 M in both HAc and NaAc, so the initial concentration of HAc and Ac- is 0.250 M.
After the addition of 0.150 mol of NaOH, it reacts with HAc to form Ac-. The stoichiometry of the reaction is 1:1, so 0.150 mol of HAc will be converted to 0.150 mol of Ac-. However, since the volume is constant, both concentrations will be affected equally.
So, the final concentration of HAc will be 0.250 M - 0.150 M = 0.100 M.
The final concentration of Ac- will be 0.250 M + 0.150 M = 0.400 M.
Now, let's calculate the pKa of HAc. The given Ka value is 1.8 x 10^-5. To convert it to pKa, take the negative logarithm:
pKa = -log(Ka)
= -log(1.8 x 10^-5)
= 4.74
Now we can plug in the values into the Henderson-Hasselbalch equation:
pH = 4.74 + log(0.400 M/0.100 M)
pH = 4.74 + log(4)
pH = 4.74 + 0.60
pH = 5.34
So, the pH of the solution after the addition of 0.150 mol of solid NaOH is approximately 5.34.