What is the pH of the solution created by combining 11.10 mL of the 0.10 M NaOH(aq)
with 8.00 mL of the 0.10 M HC2H3O2(aq)?
ok Dr. Bob so here's the work
NaOh: .00111
Acetic Acid: .0008
Salt produced: .00111-.0008= .00031
Concentration of salt: .00031/.0191= .01623
Acetic Acid unreacted: .0008-.00031= .00049
Concentration Acetic Acid: .00049/ .01091= .02565
pH= 4.74- log (.01623/.00049)= 3.22
3.22 is wrong so i tried
14-3.22= 10.78 but that's wrong too
I think part of your problem with these problems is that you aren't writing an equation and following through with an ICE chart. I'll show you how to do it even though I can't make spacing on the boards. But I can made a modified chart. It's easier to see if you write the ICE chart directly under each of the compounds in the equation.
What is the pH of the solution created by combining 11.10 mL of the 0.10 M NaOH(aq)
with 8.00 mL of the 0.10 M HC2H3O2(aq)?
ok Dr. Bob so here's the work
NaOh: .00111 This is OK.
Acetic Acid: .0008 This is OK.
Now write the equation. To keep from typing so much for acetic acid, I'm going to call it HAc.
NaOH + HAc ==> NaAc + H2O
I = initial
C = change
E = equilibrium
initial concns:
NaOH = 0.00111 mols
HAc = 0.0008 mols
NaAc = 0
H2O = 0
Change in concns:
NaOH is in excess; therefore, all of the 0.0008 mols HAc will react.
NaOH = -0.0008 mols
HAc = -0.0008 mols
NaAc = +0.0008 mols
H2O = + 0.0008 mols
Equilibrium concns:
Just add initial and change to find E.
NaOH = 0.00111-0.0008 = ??
HAc = 0.0008 - 0.0008 = 0
NaAc = 0 + 0.0008 = +0.0008
H2O = 0 + 0.0008 = +0.0008
Now you can see what you have. You have mols NaOH left unreacted, no HAc, and some salt. This is NOT a buffer. You have an excess of a strong base, NaOH, in some water and some salt. That's all there is to it.
So (NaOH) = (OH^-) = mols/L
pOH = -log(OH^-)
pH = 14 = pOH. I did a quickie calcn and got 12.21 for the pH but PLEASE check my work.
Salt produced: .00111-.0008= .00031
Concentration of salt: .00031/.0191= .01623
Acetic Acid unreacted: .0008-.00031= .00049
Concentration Acetic Acid: .00049/ .01091= .02565
pH= 4.74- log (.01623/.00049)= 3.22
3.22 is wrong so i tried
14-3.22= 10.78 but that's wrong too
Responses
The bold didn't work every where I wanted it to work but I think you can separate my work from yours.
oh my we got that answer last night too i didn't know it would be the same thing as HCl or else i wouldn't have asked. i just tried to do what we did for the acetic acid last night. but i got it now and thanks to what you taught me about the dilution stuff i have the rest of the problem right to. Thank you very much.
You're welcome.
To calculate the pH of the solution, you need to consider the dissociation of the weak acid (acetic acid, HC2H3O2) and the strong base (sodium hydroxide, NaOH). Here's the step-by-step guide to solving this problem:
1. Calculate the moles of NaOH:
Moles of NaOH = volume (in L) * concentration (in mol/L)
Moles of NaOH = 0.01110 L * 0.10 mol/L = 0.00111 mol
2. Calculate the moles of HC2H3O2:
Moles of HC2H3O2 = volume (in L) * concentration (in mol/L)
Moles of HC2H3O2 = 0.00800 L * 0.10 mol/L = 0.00080 mol
3. Determine the limiting reactant. Since NaOH and HC2H3O2 react in a 1:1 stoichiometric ratio, the limiting reactant is the one with fewer moles. In this case, HC2H3O2 is in excess, so only 0.00080 mol of it will react.
4. Calculate the moles of the salt (sodium acetate, NaC2H3O2) formed:
Moles of NaC2H3O2 = Moles of NaOH - Moles of HC2H3O2
Moles of NaC2H3O2 = 0.00111 mol - 0.00080 mol = 0.00031 mol
5. Calculate the concentration of the salt:
Concentration of NaC2H3O2 = Moles of NaC2H3O2 / Total volume (in L)
Total volume = volume NaOH + volume HC2H3O2 = 0.01110 L + 0.00800 L = 0.01910 L
Concentration of NaC2H3O2 = 0.00031 mol / 0.01910 L = 0.01623 M
6. Calculate the concentration of the unreacted HC2H3O2:
Concentration of unreacted HC2H3O2 = Moles of HC2H3O2 / Total volume (in L)
Concentration of unreacted HC2H3O2 = 0.00080 mol / 0.01910 L = 0.04178 M
7. Use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([salt] / [acid])
The pKa of acetic acid is 4.74.
pH = 4.74 + log(0.01623 / 0.04178) = 4.74 + log(0.3889) = 4.74 - 0.41 = 4.33
Therefore, the pH of the solution created by combining 11.10 mL of 0.10 M NaOH with 8.00 mL of 0.10 M HC2H3O2 is approximately 4.33.