What is the pH of the solution created by combining 2.30 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)?
mL NaOH pH wHCl pH wHC2H3O2
2.30 ________ _________
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
mL NaOH pH wHCl pH wHC2H3O2
2.30 _______ ______
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Also for the previous question that i asked, for the second part, ..
How would the volume of base change from problem three if the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water ? the volume will increase? or stay the same?
Is this two problems? or one?
one with NaOH + HCl
the other one NaOH + acetic acid.
its two sorry the whole thing that i wrote didn't show up again.
What is the pH of the solution created by combining 2.30 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)?
Here is the chart that i am supposed fill out:
mL NaOH pH w/ HCl pH w/ HC2H3O2
2.30 _________ ____________
Here's the second part of the problem.
I need to complete the chart too
mL NaOH pH w/ HCl pH w/ HC2H3O2
2.30 _________ _____________
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For my previous question, for the second part: How would the volume of base change from problem three if the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water ?
Answer: The volume of base does not change right?
The NaOH + HCl part you know how to do from the previous problem I worked. And you are right about the volume of base not changing if the solution is diluted AFTER putting it in the titration vessel.
For the NaOH + HC2H3O2 problem, you need to write the equation. It is
NaOH + HC2H3O2 ==> NaC2H3O2 + HOH.
Determine mols NaOH from mols = M x L.
Determine mols HC2H3O2 from M x L = ??
This problem will not be the same kind as the NaOH + HCl BECAUSE in NaOH/HCl you are titrating a strong acid with a strong base and both are ionized 100%. With NaOH + acetic acid, you must realize that this is the addition of a strong base to a weak acid. I haven't gone through the numbers but I'm sure acetic acid is in excess which means you are left with some sodium acetate and some unreacted acetic acid. That is a buffer. Solve that with the Henderson-Hasselbalch equation. I'll let you worry about the chart.
Post your work if you need additional help.
me thinks u right
To find the pH of a solution, we need to understand how the reactants (acids and bases) interact and determine the concentration of hydrogen ions in the final solution.
For the first part of the question:
To calculate the pH of the solution created by combining 2.30 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HCl(aq), we can use the concept of neutralization reaction. In this reaction, the acid (HCl) reacts with the base (NaOH) to form water and a salt.
The balanced chemical equation for the neutralization reaction is:
HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)
Since the neutralization reaction is stoichiometric, we can determine the moles of HCl and NaOH that react based on their initial concentrations and volumes.
For HCl:
Moles of HCl = (0.10 mol/L) * (8.00 mL / 1000 mL/mL) = 0.008 mol
For NaOH:
Moles of NaOH = (0.10 mol/L) * (2.30 mL / 1000 mL/mL) = 0.0023 mol
Considering that the reaction is stoichiometric, all the HCl and NaOH molecules react to form water and NaCl. No excess of either reactant should remain after the reaction.
Therefore, we can find the concentration of the resulting H+ ions in solution. Since 1 mole of HCl produces 1 mole of H+ ions, and 1 mole of NaOH produces 1 mole of OH- ions, the concentration of H+ ions in the final solution will depend on the excess of reactant.
In this case, HCl is the excess reactant because the moles of HCl (0.008 mol) are greater than the moles of NaOH (0.0023 mol).
To find the concentration of H+ ions in the final solution, we consider the excess moles of HCl after the reaction. Subtracting the moles of NaOH from the moles of HCl gives the excess moles of HCl:
Excess moles of HCl = 0.008 mol - 0.0023 mol = 0.0057 mol
Since the reaction is stoichiometric, 0.0057 mol of HCl will produce 0.0057 mol of H+ ions.
Now we need to calculate the volume of the final solution. In this case, we are given the volumes of NaOH and HCl. To get the total volume, we sum the volumes of NaOH and HCl:
Total volume = volume of NaOH + volume of HCl = 2.30 mL + 8.00 mL = 10.30 mL
Since the volume of the final solution is 10.30 mL, we need to convert it to liters:
Total volume = 10.30 mL = 10.30 mL * (1 L / 1000 mL) = 0.01030 L
Finally, we can calculate the concentration of H+ ions in the final solution (pH) using the formula for pH:
pH = -log[H+]
pH = -log(0.0057 mol / 0.01030 L) = -log(0.554) = 0.256
Therefore, the pH of the solution created by combining 2.30 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HCl(aq) is approximately pH 0.256.
Now, moving on to the second part of the question:
If the 8.00 mL of 0.10 M acid (HCl) was first diluted with 100 mL of water, the volume of the acid will increase, but the moles of the acid will remain the same. This is because dilution involves adding a solvent (water) to decrease the concentration of the solute (acid), while keeping the number of moles constant.
So, the volume of the base (NaOH) will remain the same, i.e., 2.30 mL, as the volume and concentration of NaOH are independent of the dilution of the acid.