Asked by Ethio
An unbuffered solution contaning 0.000018 M HCl .Calculate PH after adding 1 mL of 0.1 M NaOH to 100 mL of solution.
Answers
Answered by
DrBob222
millimoles HCl = mL x M = 100 x 0.000018M = 0.0018.
millimoles NaOH added = 1 mL x 0.1M = 0.1.
..........HCl + NaOH ==> NaCl + H2O
initial..0.0018..0.1.......0......0
change..-0.0018.-0.0018.+0.0018.0.0018
equil....0......0.0982...0.0018.0.0018
You can see that all of the HCl was neutralized and one has a solution of NaCl in excess NaOH.
(OH^-) = 0.0982 mmoles/101 mL = 0.000972M
pOH = -log(OH^-) = -log(0.0.000972 = 3.012
Then pH + pOH = pKw = 1E-14
Solve for pH. It's about 11.
Check my numbers. I've been bad about typos today, especially with the zeros (adding them or omitting them)
millimoles NaOH added = 1 mL x 0.1M = 0.1.
..........HCl + NaOH ==> NaCl + H2O
initial..0.0018..0.1.......0......0
change..-0.0018.-0.0018.+0.0018.0.0018
equil....0......0.0982...0.0018.0.0018
You can see that all of the HCl was neutralized and one has a solution of NaCl in excess NaOH.
(OH^-) = 0.0982 mmoles/101 mL = 0.000972M
pOH = -log(OH^-) = -log(0.0.000972 = 3.012
Then pH + pOH = pKw = 1E-14
Solve for pH. It's about 11.
Check my numbers. I've been bad about typos today, especially with the zeros (adding them or omitting them)
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MULAT
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