Asked by Mizuhara
A 25 mL sample of 0.29 M LiOH analyte was titrated with 0.75 M HI at 25 degree Celsius. 1)Calculate the initial pH before any titrant was added. 2) Then calculate the pH of the solution after 5 mL of titrant was added.
Answers
Answered by
DrBob222
At the beginning you have 0.29 M LiOH. That ionizes 100% so (OH^-) = 0.29M
pOH = -log(OH^-); solve for pOH,then
pOH + pH = pKw = 14. YOu know pKw and pOH solve for pH.
To start we have 25 mL x 0.29 M = approx 7 millimols LiOH but you need a better answer on this as well as all of the other calculations that follow.
For pH after the addition of 5.0 mL of 0.75M that is 3.75 mmols HI added.
..............LiOH + HI ==> H2O + LiI
I.............7.....0......0.....0
add................3.75
C...........-3.75.-3.75.....3.75..3.75
E......approx3.25..0........3.75..3.75
M LiOH remaining is mmols/mL = approx 3.25/(25+5) = ?
Then convert to pOH followed by convert to pH.
pOH = -log(OH^-); solve for pOH,then
pOH + pH = pKw = 14. YOu know pKw and pOH solve for pH.
To start we have 25 mL x 0.29 M = approx 7 millimols LiOH but you need a better answer on this as well as all of the other calculations that follow.
For pH after the addition of 5.0 mL of 0.75M that is 3.75 mmols HI added.
..............LiOH + HI ==> H2O + LiI
I.............7.....0......0.....0
add................3.75
C...........-3.75.-3.75.....3.75..3.75
E......approx3.25..0........3.75..3.75
M LiOH remaining is mmols/mL = approx 3.25/(25+5) = ?
Then convert to pOH followed by convert to pH.
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