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Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula. What is the pH of the solution...Asked by Sara
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)?
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)?
Answers
Answered by
DrBob222
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.
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.
Answered by
Billy
im having trouble with the samee one....
i got 1.72 for the pH and its still wrong
i got 1.72 for the pH and its still wrong
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