Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
What is the pH of the solution created by combining 11.80 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)?
11.80 mL x 0.1M NaOH = 1.180 mmoles.
8.00 mL x 0.1M HCl = 0.800 mmoles
8.00 mL x 0.1M HAc = 0.800 mmoles.
............HCl + NaOH ==> NaCl + H2O
initial..0.800...1.180.....0.......0
change..-0.800..-0.800.....0.800.0.800
equi.......0.....0.380.....0.800..0.800
So you have 0.380 mmoles NaOH in 11.8+8.00 mL soln and (OH^-) = (NaOH) and convert to pH.
..........HAc + NaOH ==> NaAc + H2O
initial..0.800..1.180.....0......0
change..-0.800..-0.800..0.800..0.800
equil....0......0.380...0.800..0.800
Ignoring the OH^- that might be added due to the hydrolysis of the NaAc salt (which is negligible) pH is determined by the mmoles NaOH/mL.
The second half is all yours. Post your work if you get stuck.
im having trouble with the samee one....
i got 1.72 for the pH and its still wrong
To find the pH of the solution created by combining the given solutions, we need to consider the stoichiometry and the dissociation of the acid and base.
First, let's calculate the moles of NaOH and HCl added to the solution. We can use the formula:
moles = concentration (in M) x volume (in L)
For NaOH:
moles of NaOH = 0.10 M x 0.01180 L = 0.00118 mol
For HCl:
moles of HCl = 0.10 M x 0.00800 L = 0.00080 mol
Now, let's consider the stoichiometry between the acid (HCl) and base (NaOH). Since HCl is a strong acid and NaOH is a strong base, they react in a 1:1 ratio according to the balanced equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Since the moles of HCl and NaOH are not equal (0.00080 mol vs. 0.00118 mol), we will use the limiting reagent to determine the moles of the remaining reactant.
Since HCl is the limiting reagent, all of it will react with NaOH, leaving no excess HCl. Therefore, after the reaction, we have used all 0.00080 mol of HCl and still have 0.00118 - 0.00080 = 0.00038 mol of NaOH remaining.
Now, let's calculate the new concentration of NaOH in the solution.
volume of solution = volume of NaOH + volume of HCl = 0.01180 L + 0.00800 L = 0.01980 L
new concentration of NaOH = moles of NaOH remaining / volume of solution
= 0.00038 mol / 0.01980 L
= 0.0192 M
To find the pH, we need to calculate the pOH first using the concentration of NaOH. The pOH can be calculated using the formula:
pOH = -log[OH-]
Since NaOH is a strong base, it fully dissociates into Na+ and OH-. Therefore, [OH-] is equal to the concentration of NaOH itself.
pOH = -log(0.0192)
≈ 1.72
Finally, to get the pH, we can use the equation:
pH = 14 - pOH
= 14 - 1.72
≈ 12.28
Therefore, the pH of the solution created by combining 11.80 mL of 0.10 M NaOH with 8.00 mL of 0.10 M HCl is approximately 12.28.
To calculate the pH values when the 8.00 mL of 0.10 M acid is diluted with 100 mL of water, we need to consider the dilution effect.
Step 1: Calculate the moles of the acid originally in 8.00 mL:
moles of acid = 0.10 M x 0.00800 L = 0.00080 mol
Step 2: Calculate the moles of the acid after dilution:
moles of acid after dilution = moles of acid / dilution factor
The dilution factor is the ratio of final volume to initial volume. In this case, the final volume is 8.00 mL + 100 mL = 108.00 mL.
dilution factor = final volume / initial volume = 108.00 mL / 8.00 mL = 13.5
moles of acid after dilution = 0.00080 mol / 13.5 ≈ 0.000059 mol
Step 3: Calculate the new concentration of the acid after dilution:
new concentration of acid = moles of acid after dilution / final volume
final volume = 108.00 mL / 1000 mL/L = 0.10800 L
new concentration of acid = 0.000059 mol / 0.10800 L ≈ 0.00055 M
Step 4: Calculate the pOH using the new concentration of the acid just like in the previous example.
Finally, subtract the pOH from 14 to get the pH value.
This way, you can find the pH values when taking into account the dilution of the acid solution with water.