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.
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.
To calculate the pH of the buffer solution after the addition of NaOH, we need to consider the reaction that takes place between HC2H3O2 and NaOH.
The reaction can be expressed as follows:
HC2H3O2 + NaOH → NaC2H3O2 + H2O
Since the given buffer solution is 0.250 M in both HC2H3O2 and NaC2H3O2, the initial moles of HC2H3O2 and NaC2H3O2 can be calculated as follows:
moles of HC2H3O2 = (0.250 M) x (1.00 L) = 0.250 mol
moles of NaC2H3O2 = (0.250 M) x (1.00 L) = 0.250 mol
When 0.150 mol of solid NaOH is added, it reacts completely with HC2H3O2 in a 1:1 ratio.
After the reaction, the moles of HC2H3O2 will decrease by 0.150 mol, and the moles of NaC2H3O2 will increase by 0.150 mol.
The remaining moles of HC2H3O2 can be calculated as follows:
remaining moles of HC2H3O2 = initial moles of HC2H3O2 - moles of NaOH added = 0.250 mol - 0.150 mol = 0.100 mol
Next, we can calculate the moles of NaC2H3O2:
moles of NaC2H3O2 = initial moles of NaC2H3O2 + moles of NaOH added = 0.250 mol + 0.150 mol = 0.400 mol
Now, we can calculate the concentration of HC2H3O2 and NaC2H3O2 after the reaction:
concentration of HC2H3O2 = remaining moles of HC2H3O2 / (volume of solution in liters) = 0.100 mol / 1.00 L = 0.100 M
concentration of NaC2H3O2 = moles of NaC2H3O2 / (volume of solution in liters) = 0.400 mol / 1.00 L = 0.400 M
Since HC2H3O2 is a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:
pH = pKa + log ([A-] / [HA])
where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base (in this case, NaC2H3O2), and [HA] is the concentration of the acid (in this case, HC2H3O2).
Given that the Ka value for HC2H3O2 is 1.8 x 10^-5, the pKa can be calculated as follows:
pKa = -log (1.8 x 10^-5) = 4.74
Plugging in the values, the pH can be calculated as follows:
pH = 4.74 + log (0.400 M / 0.100 M)
pH = 4.74 + log (4.00)
Using a calculator, we find that log (4.00) is approximately 0.6021.
Therefore, the pH of the buffer solution after the addition of 0.150 mol of solid NaOH is approximately 5.3421.