What is the pH of the solution created by combining 2.40 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HC2H3O2(aq)?
Can somebody explain how to do this? I am stuck. I've tried solving for H+ concentration. I got .00024M and subtracted it from [OH-]. Then I solved for pOH and used that to get pH but still got the wrong answer. Somone plz explain this to me.
I think it's a simple problem IF you recognize what you have. To simplify, let's call HC2H3O3, acetic acid, HAC. Then
NaOH + HAc ==> NaAc + H2O
moles NaOH initially = L x M = 0.00240 x 0.1 = 0.000240
moles HAc = L x M = 0.008 x 0.1 M = 0.0008.
Note the NaOH is the smaller amount; therefore, HAc will neutralize all of the NaOH and some HAc will be left over.
Final mols.
NaOH = 0
HAc = 0.0008-0.00024 = 0.00056
NaAc formed = 0.00056
H2O formed but not worry with that.
So what do you have when all of this is over? We have a weak acid (HAc) and its salt (NaAc) which makes a buffer. Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log[(base)/(acid)]
base = concn of Ac^- which is moles/L.
acid = concn of HAc which is moles/L.
Plug and chug. I'll post this, then repost with an approximate answer to help you check your answer.
You should get close to 4.3 or so for pH.
How do you find the concentration of the conjugate base?
base = mols acetate/L
moles acetate = 0.00056
volume = 10.4 mL = 0.0104 L.
concn = 0.00056/0.0104 = ??
How are we finding pKa for the Henderson-Hasselbalch equation?
Never-mind decided just to use a table of pKa values. By the way the NaAc formed wouldn't be 0.00056 it would be 0.00024 because you can only make as much product as your limiting reagent allows (in this case OH- ions) I seriously tried using these numbers a ton of times and it was the NaAc value that screwed everything up.
To find the pH of the solution, we can follow these steps:
Step 1: Calculate the moles of NaOH and HC2H3O2 used.
Moles NaOH = (Volume NaOH × Concentration NaOH) / 1000
= (2.40 mL × 0.10 M) / 1000
= 0.00024 moles of NaOH
Moles HC2H3O2 = (Volume HC2H3O2 × Concentration HC2H3O2) / 1000
= (8.00 mL × 0.10 M) / 1000
= 0.0008 moles of HC2H3O2
Step 2: Determine the limiting reactant. The limiting reactant is the one that produces fewer moles of H+ ions. In this case, we have 1 mole of NaOH reacts with 1 mole of HC2H3O2 to produce 1 mole of H+ ions. Since the moles of NaOH and HC2H3O2 used are not in a 1:1 ratio, we need to determine which one is the limiting reactant.
To do this, we compare the moles of the two reactants. The reactant with fewer moles is the limiting reactant. In this case, both NaOH and HC2H3O2 have fewer moles of H+ ions. Therefore, neither of them is a limiting reactant.
Step 3: Calculate the total volume of the mixed solution.
Volume_total = Volume NaOH + Volume HC2H3O2
= 2.40 mL + 8.00 mL
= 10.40 mL (or 0.0104 L)
Step 4: Calculate the total moles of H+ ions produced.
Total moles H+ ions = Moles NaOH + Moles HC2H3O2
= 0.00024 moles of NaOH + 0.0008 moles of HC2H3O2
= 0.00104 moles of H+ ions
Step 5: Determine the concentration of H+ ions in the solution.
Concentration H+ ions = Total moles H+ ions / Volume_total
= 0.00104 moles / 0.0104 L
= 0.1 M
Step 6: Calculate the pOH of the solution.
pOH = -log[OH-]
= -log(0.1)
Step 7: Calculate the pH of the solution.
pH = 14 - pOH
By following these steps, you should be able to calculate the pH of the solution. Make sure to double-check your calculations and units to ensure accuracy.