What is the pH obtained by mixing 50.00mL of 0.100N HOAc (acetic acid) and 25.00mL of 0.200N NaOH?
Use the Henderson-Hasselbalch equation.
Alright I have the equation but I still don't get how to set it up.
To find the pH obtained by mixing 50.00 mL of 0.100N HOAc (acetic acid) and 25.00 mL of 0.200N NaOH, we need to calculate the concentration of the resulting solution.
The balanced chemical equation for the reaction between HOAc (acetic acid) and NaOH is:
HOAc + NaOH -> NaOAc + H2O
From the equation, we can see that 1 mole of HOAc reacts with 1 mole of NaOH to produce 1 mole of NaOAc (sodium acetate) and 1 mole of water.
The number of moles of HOAc can be calculated using the formula:
moles = concentration (M) x volume (L)
For HOAc:
moles of HOAc = 0.100N x 0.05000 L = 0.005 moles
Similarly, the number of moles of NaOH can be calculated:
moles of NaOH = 0.200N x 0.02500 L = 0.005 moles
As per the balanced equation, since the mole ratio between HOAc and NaOH is 1:1, both react completely, and we have 0.005 moles of NaOAc produced in the reaction.
To find the concentration of the resulting solution, we need to calculate the total volume:
total volume = volume of HOAc + volume of NaOH = 50.00 mL + 25.00 mL = 75.00 mL = 0.0750 L
The concentration of the resulting solution is then:
concentration (N) = moles of NaOAc / total volume
concentration (N) = 0.005 moles / 0.0750 L = 0.067 N
Now, to find the pH of the resulting solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, HOAc acts as an acid, and NaOAc acts as its conjugate base. The pKa value for acetic acid is 4.76. We can assume that the number of moles of NaOAc is equal to the number of moles of HOAc. So, [A-]/[HA] = 1.
Now, substituting the values into the equation:
pH = 4.76 + log(1)
pH = 4.76
Therefore, the pH obtained by mixing 50.00 mL of 0.100N HOAc and 25.00 mL of 0.200N NaOH is 4.76.