Question 6.
Part (a).
The pH of 10-1 M acetic acid = 1/2 (pKa - Log[acetic acid]) = 1/2 (4.74 - Log10-1) = 2.87
Since 10-3 M basic salt (Na2CO3) is added, the pH will increase slightly.
Therefore, the pH of the resulting solution ~ 2.97
Part (b).
If 10-3 M NaOH is added to the solution in Part (a), the pH will further increase.
Therefore, the pH of the resulting solution ~ 3.07
6. (a) Construct a log Il-pH diagram to determine the pH of a solution containing 10-3...
I need help with just the last part where it asks to find the
pH.
A solution containing 0.1 7 M Pb". 1.5 x 10-6 М РЬ". 1.5 x 10-6 M Mn". 0.17 M MnO-, and 0.88 M HNO, was prepared. For this solution, the balanced reduction half-reactions and overall net reaction shown can occur. E1.690Vv E1.507 V A. Determine E AG, and K for this reaction. Eel0.183 AG-176600 9.024 x1030 B. Calculate the value for the cell potential, Ecell,...
A solution containing 0.190.19 M Pb2+Pb2+ , 1.5×10−61.5×10−6 M Pb4+Pb4+ , 1.5×10−61.5×10−6 M Mn2+Mn2+ , 0.190.19 M MnO−4MnO4− , and 0.900.90 M HNO3HNO3 was prepared. For this solution, the balanced reduction half‑reactions and overall net reaction shown can occur. 5[Pb4++2e−↽−−⇀Pb2+]5[Pb4++2e−↽−−⇀Pb2+] ?∘+=1.690 VE+°=1.690 V 2[MnO−4+8H++5e−↽−−⇀Mn2++4H2O]2[MnO4−+8H++5e−↽−−⇀Mn2++4H2O] ?∘−=1.507 VE−°=1.507 V 5Pb4++2Mn2++8H2O↽−−⇀5Pb2++2MnO−4+16H+5Pb4++2Mn2++8H2O↽−−⇀5Pb2++2MnO4−+16H+ A. Determine ?∘cellEcell∘, Δ?∘ΔG∘, and ?K for this reaction. ?∘cell=Ecell∘= V Δ?∘=ΔG∘= J ?=K= B. Calculate the value for the cell potential, ?cellEcell, and the free energy, Δ?ΔG, for the given conditions....
Titration: Acids and Bases
2. How can you determine which acid is diprotic?
3. using the answers to questions one and two, which acid is
diprotic?
4. Which base has more hydroxide ions per molecule?
Acid Volume Base Base Initial Volume (mL) Base Final Volume (mL) Volume of Base Used (mL) Acid: Base Ratio Acid 1 20 mL Base 1 50 mL 34.5 15.5 4:3 Acid 2 20 mL Base 1 Acid 1 20 mL Base 2 Acid 2 20...