Asked by Sarah
Determine the pH of a solution after 20.00 mL of 0.4963 M HI has been titrated with 12.64 mL of 0.5174 M NaOH.
Answers
Answered by
DrBob222
millimols HI = mL x M = about 9.9 but you need a better answer on this AND all that follows. My numbers are just estimates.
mmols NaOH = approx 6.5
......HI + NaOH ==> NaI + H2O
I....9.9...0.........0.....0
add........6.5...............
C...-6.5..-6.5......6.5...6.5
E... 3.4...0........6.5...6.5
So the solution at the end is 3.4 mmols HI and 6.5 mmols NaI. HI is a strong acid and NaI is neutral in water; therefore, the pH is determined by the HI.
M HI = mmols/mL = ?
Convert that to pH.
Remember to do this more accurately from the beginning. Watch the significant figures.
mmols NaOH = approx 6.5
......HI + NaOH ==> NaI + H2O
I....9.9...0.........0.....0
add........6.5...............
C...-6.5..-6.5......6.5...6.5
E... 3.4...0........6.5...6.5
So the solution at the end is 3.4 mmols HI and 6.5 mmols NaI. HI is a strong acid and NaI is neutral in water; therefore, the pH is determined by the HI.
M HI = mmols/mL = ?
Convert that to pH.
Remember to do this more accurately from the beginning. Watch the significant figures.
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