What is the pH of the solution created by combining 11.40 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?
with 8.00 mL of the 0.10 M HC2H3O2(aq)?
Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
Table 1)
mL NaOH pHw/HCl pH w/HC2H3O2
11.40 ______ _______
Table 2)
mL NaOH pHw/HCl pH w/HC2H3O2
11.40 ______ ________
ive done this problem literally 28 times...please help with details i really appreciate it
11.4 mL x 0.1M NaOH = 1.14 mmoles NaOH.
8.00 mL x 0.1M HCl = 0.800 mmoles HCl.
9.00 mL x 0.1M HAc = 0.800 mmoles HAc
...........NaOH + HCl ==> NaCl + H2O
initial..1.14....0.800.....0......0
change..-0.800....-0.800..0.800..0.800
equil....0.34......0......0.800..0.800
Looking at the equilibrium line, NaCl is not hydrolyzed so you have a solution of NaOH. (OH^-) = (NaOH) = mmoles/total mL. Convert to pOH then to pH.
For the acetic acid part,
...........NaOH + HAc ==> NaAc + H2O
initial....1.14...0.800....0......0
change....-0.800..-0.800..0.800..0.800
equil......0.34...0......0.800..0.800
Here the NaAc salt does hydrolyze; however, the OH^- produced by the hydrolysis of the Ac^- is so small it can be neglected so the (OH^-) = (NaOH) = mmoles NaOH/total mL. Convert to pOH then to pH.
I don't know enough about the problem to answer about the dilution part. I'll leave that for you .
To find the pH of the solution created by combining NaOH with HCl and HC2H3O2, we need to understand the concept of acid-base reactions and how they affect the pH.
1. Find the moles of NaOH, HCl, and HC2H3O2:
- NaOH: 11.40 mL * 0.10 M = 1.14 mmol
- HCl: 8.00 mL * 0.10 M = 0.80 mmol
- HC2H3O2: 8.00 mL * 0.10 M = 0.80 mmol
2. Determine the limiting reagent:
- The reactant that runs out first is the limiting reagent.
- In this case, since both HCl and HC2H3O2 have the same number of moles, they will both be completely consumed, making them the limiting reagent.
3. Calculate the moles of NaOH remaining:
- NaOH is in excess, so it will not be completely consumed.
- The difference in moles = Initial moles of NaOH - Moles of HCl or HC2H3O2
- Moles of NaOH remaining = 1.14 mmol - 0.80 mmol = 0.34 mmol
4. Convert the remaining NaOH moles to concentration:
- Concentration of NaOH = Moles of NaOH remaining / Total volume of solution (11.40 mL + 8.00 mL)
- Concentration of NaOH = (0.34 mmol / 19.40 mL) * (1000 mL / 1 L) = 17.53 mM
5. Calculate the pOH of the solution:
- pOH = -log10(OH- concentration)
- OH- concentration = NaOH concentration
- pOH = -log10(17.53 mM) = 1.757
Now, let's complete the tables based on the calculations:
Table 1)
mL NaOH pHw/HCl pHw/HC2H3O2
11.40 1.757 (To be calculated)
Table 2)
mL NaOH pHw/HCl pHw/HC2H3O2
11.40 1.757 (To be calculated)
To find the pH with HC2H3O2, we need to consider that the 8.00 mL of 0.10 M acid was diluted with 100 mL of water. However, the dilution does not affect the concentration of the acid because the number of moles remains the same.
6. Repeat steps 2-5 using the diluted volume of the acid:
- Moles of HC2H3O2 = 8.00 mL * 0.10 M = 0.80 mmol
- Moles of NaOH remaining = 1.14 mmol - 0.80 mmol = 0.34 mmol
- Concentration of NaOH = (0.34 mmol / 119.40 mL) * (1000 mL / 1 L) = 2.84 mM (Note: total volume is 119.40 mL)
- pOH = -log10(2.84 mM) = 2.546
Now, let's complete the tables based on the updated calculations:
Table 1)
mL NaOH pHw/HCl pHw/HC2H3O2
11.40 1.757 2.546
Table 2)
mL NaOH pHw/HCl pHw/HC2H3O2
11.40 1.757 2.546
Therefore, the pH values for the two cases are 1.757 for the solution with HCl and 2.546 for the solution with HC2H3O2.