What is the pH of the solution created by combining 11.50 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)?
What are the pH values for HCl and HC2H302 if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water?
HCl + NaOH ==> NaCl + HOH
1:1 ratio strong acid vs stong base.
mols = M x L and that's all you need.
Determine mols NaOH which will equal mols HCl and M = mols/L for HCl.
For acetic acid. This is a strong base/weak acid. It forms a buffer solution. Use the Henderson-Hasselbalch equation.
Write the equation.
Calculate mols NaOH.
Calculate mols acetic acid.
Determine how much sodium acetate is formed. Determine how much acetic acid remains unreacted. The excess acetic acid and the sodium acetate salt formed is the buffer. The H-H equation will give you the pH.
Post your work if you get stuck.
(The last part is just the set of problems over again but diluted first. I assume you know how to convert H^+ to pH>)
ok for a lab we had to titrate a solution with KMnO4 and for each trial i got 22.83 mL,23.92, and 22.48
I need to find the mass percent of the oxalate ion. How would I do this?
The equation is MnO4-+C2O4 2- -> Mn 2+ + CO2
In the lab we used about .1 g of the oxalate salt in each titration.
To determine the pH of a solution, we need to consider the concentration of the hydronium ions (H3O+).
For the first part of your question, we will calculate the pH of the solution formed by combining 11.50 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HCl(aq). Since NaOH is a strong base and HCl is a strong acid, they will react completely to form water and a salt. The balanced equation for the reaction is:
NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)
Since this reaction goes to completion, we can determine the concentration of the remaining ions in the solution. The solution will contain the anion from NaCl (Cl-) and the cation from NaOH (Na+), both of which will not have any effect on the pH.
Therefore, the pH of this solution will be determined by the water present, which is neutral. Thus, pH = 7.
For the second part of your question, we will calculate the pH values for HCl and HC2H3O2 after dilution by 100 mL of water.
First, we need to calculate the new concentration of the acid after dilution. The initial molarity of HCl is 0.10 M. We can calculate the new concentration using the dilution formula:
C1V1 = C2V2
where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume. In this case, C1 = 0.10 M, V1 = 8.00 mL, C2 = ?, and V2 = 8.00 mL + 100 mL.
Using the dilution formula, we can solve for C2:
0.10 M × 8.00 mL = C2 × (8.00 mL + 100 mL)
C2 = (0.10 M × 8.00 mL) / (8.00 mL + 100 mL)
C2 = 0.008 M
Now that we have the new concentration, we can calculate the pH. For HCl, it dissociates completely in water to release H3O+ ions. Therefore, the concentration of H3O+ ions is equal to the concentration of HCl.
So, the pH of the HCl solution after dilution will be determined by the concentration of H3O+ ions, which is 0.008 M. To find the pH, we use the equation pH = -log[H3O+]:
pH = -log(0.008) ≈ 2.10
Now, for HC2H3O2, it is a weak acid that does not dissociate completely in water. However, after dilution, we can assume complete dissociation since the concentration is relatively low.
The equation for the dissociation of HC2H3O2 is:
HC2H3O2(aq) + H2O(l) ↔ H3O+(aq) + C2H3O2-(aq)
Here, HC2H3O2 acts as an acid, donating an H+ ion to the water to form H3O+. Therefore, the concentration of H3O+ ions is equal to the concentration of HC2H3O2.
The pH of the HC2H3O2 solution after dilution will be determined by the concentration of H3O+ ions, which is 0.008 M. Using the equation pH = -log[H3O+]:
pH = -log(0.008) ≈ 2.10
So, the pH values for both HCl and HC2H3O2 after dilution with 100 mL of water will be approximately 2.10.