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>)
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?
2 answers
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.
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.