Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Gobo on March 30, 2012, 06:55:06 PM
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Assume we start with 0.025M NaOH. 100mL of this 0.025M NaOH is poured into 0.5g Ca(OH)2, and this will eventually form a saturated solution.
Then, a 25mL aliquot of this solution is taken. What is the concentration of NaOH now?
Attempt:
0.025mol/L * (0.1L) = 0.0025mol of NaOH in total
Then when you take 25mL from this 100mL part of it, we get:
0.0025mol / 0.025L = 0.1M NaOH
So, when you take 25mL portion of this solution, we have concentration of 0.1M NaOH now...?
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You calculated that you have 0.0025 mol NaOH in the 100 mL portion that was transferred to Ca(OH)2
When you removed 25 mL from this 100 mL portion, how many moles of NaOH did you remove? In other words, how many moles NaOH were present in the 25 mL removed?
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You calculated that you have 0.0025 mol NaOH in the 100 mL portion that was transferred to Ca(OH)2
When you removed 25 mL from this 100 mL portion, how many moles of NaOH did you remove? In other words, how many moles NaOH were present in the 25 mL removed?
So you are implying that instead of this:
0.025mol/L * (0.1L) = 0.0025mol of NaOH in total
It should be:
0.025mol/L * (0.025L) = 0.000625 mol NaOH?
So if I do that, I get the mol of OH- from Ca(OH)2, and use the moles of OH- from NaOH (1:1 ratio so it is the same as above) and divide this by 0.025L, and then I get the [OH-]?
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The first answer was quite correct. You did not add no water to your solution. Since NaOH does not react with calcium hydroxide its concentration remains practically unchanged (without any calculations!).
We can expect a very, very small change of volume after addition of solid Ca(OH)2, but we have no information that allows correction of volume.
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To add to what AWK wrote - you can expect a slight change of pOH, as even in the solution of OH- some small amount of Ca(OH)2 will dissolve. But NaOH is mostly a spectator here, so its concentration can't change.