What is the pH of the solution that results from mixing 5 ml of 0.2 M NaOH and 100 mL of 0.05 M lactic acid? (Pka of lactic acid= 3.86)
Responses
Use the Henderson-Hasselbalch equation. The mixture you have indicated will react to form a buffer, the base will be the sodium lactate and the acid will be the lactic acid not neutralized by the NaOH. Post your work if you get stuck.
I know that I'm supposed to use the equation:
pH= pKa+log(deprotnated, which case the lactic acid not neutralized)/(protonated, sodium lactate)
but I'm not sure where the numbers come into play. i got that initally there are 0.001 mol of NaOH and 0.005 mol of lactic acid. I tried doing limiting reagents and got that the products each get 0.001 mol? which would give the ratio of 1, which i don't think is right...
So what is the reaction? I'll call lactic acid HL for simplicity.
HL + NaOH ==> NaL + H2O
HL = 0.005 mols. You are correct.
NaOH = 0.001 mols. You are correct.
Now you see that an acid + base = salt + water.
So the smaller mols will react completely and leave an excess of what is left.
0.001 mol NaOH will form 0.001 mol NaL, 0.001 mol H2O, and will leave 0 mol NaOH (all of it is reacted), but 0.005 - 0.001 mol = 0.004 mols HL.
So you have an acid (0.004 mol HL) and a salt of the weak acid (0.001 NaL) so that is a buffered solution.
(NaL) = (0.001 mol/0.105 L)
(HL) = (0.004 mol/0.105 L)
The H-H equation is
pH = pKa + log (base/acid)
base is NaL
acid is HL.
lolol
To determine the pH of the solution, you will need to use the Henderson-Hasselbalch equation. This equation relates the pH of a solution to the pKa value of the acid and the ratio of its conjugate base to its acid form.
In this case, you have 5 mL of 0.2 M NaOH and 100 mL of 0.05 M lactic acid. To use the Henderson-Hasselbalch equation, you first need to calculate the moles of acid and base present in the solution.
For the NaOH solution, you have 5 mL (which can be converted to L by dividing by 1000) and a concentration of 0.2 M. Therefore, the moles of NaOH can be calculated as follows:
moles of NaOH = volume (L) x concentration (M)
= (5 mL / 1000) L x 0.2 M
= 0.001 mol NaOH
For the lactic acid solution, you have 100 mL (also converted to L) and a concentration of 0.05 M. Thus, the moles of lactic acid can be calculated as:
moles of lactic acid = volume (L) x concentration (M)
= (100 mL / 1000) L x 0.05 M
= 0.005 mol lactic acid
Now that you have the moles of NaOH and lactic acid, you need to determine the ratio of their conjugate base to their acid form. Since NaOH is a strong base, it will completely react with lactic acid to form its conjugate base, sodium lactate. This means that the moles of sodium lactate formed will be equal to the moles of NaOH used.
Therefore, you have 0.001 mol of sodium lactate and 0.005 mol of lactic acid.
Now, you can substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([sodium lactate] / [lactic acid])
The pKa of lactic acid is provided as 3.86.
Substituting the values:
pH = 3.86 + log(0.001 / 0.005)
Simplifying the expression:
pH = 3.86 + log(0.2)
Using a logarithm calculator, you can find the value of log(0.2) to be -0.699.
Therefore,
pH = 3.86 - 0.699
= 3.161
So, the pH of the solution resulting from mixing 5 mL of 0.2 M NaOH and 100 mL of 0.05 M lactic acid is approximately 3.161.