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Important note: I'm an eighth-grader. I taught myself chemistry to organic chem using an arsenal of textbooks*, but I have gotten things wrong, namely resonance.


Please read above (also this if you want more background on the problems).

You have been given a sample of Crystal Drano®. There are two components in the Drano – some small shiny metallic pieces, and some pale green beads. (The green color is a dessicating substance.) The metallic pieces are either zinc, magnesium, or aluminum. The beads are either $\ce{NaOH}$, $\ce{Ca(OH)2}$, or $\ce{Al(OH)3}$.

I recognized this as a titration problem immediately. I dissolved $\pu{1.6 grams}$ of the green beads in $\pu{50 mL}$ of distilled water. I filtered out the metal and added a total of $\pu{6 mL}$ of $\pu{3M }\ce{HCl}$ into the solution, causing the indicator to change color. I added $\pu{1.75 mL}$ of $\pu{1 M NaOH}$ using a more precise pipette (it doesn't say that you can use a buret) to titrate it completely.

Then I calculated these values (molar mass):

$\ce{NaOH} = \pu{40 g/mol}$

$\ce{Ca(OH)2} = \pu{74 g/mol}$

$\ce{Al(OH)3} = \pu{78 g/mol}$

$\pu{Mystery} = \pu{(\frac{3M}{L} * \frac{1L}{1000mL} * 6mL) - (\frac{1M}{L} * \frac{1L}{1000mL} * 1.75mL) = 13/1000 M in 1.6 g \approx 48 g/mol} $

The mystery material(still) seems to be off by a sizeable amount. Why? All I can think about is that maybe the base reacted with the metal in aqueous solution to create a metal salt.

*Not impressive, only about 7-10 textbooks relying mostly on Atkins

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  • $\begingroup$ I don't understand very well what you did. Do the $\pu{1.6 g}$ are just for the beads? Or does that include the metal as well? Also, the molar mass of $\ce{NaOH}$ is $\pu{40 g mol-1}$. $\endgroup$ – ralk912 Mar 27 '18 at 21:19
  • $\begingroup$ If your titration and values are correct, then $\pu{16.25 mmol}$ of $\ce{HCl}$ reacted with the bead compound. This would mean that you initially had either $\pu{650 mg}$ of $\ce{NaOH}$, $\pu{1.20 g}$ of $\ce{Ca(OH)2}$, or $\pu{1.27 g}$ of $\ce{Al(OH)3}$. $\endgroup$ – ralk912 Mar 27 '18 at 21:29
  • $\begingroup$ Oh shoot. Have to edit... $\endgroup$ – JavaScriptCoder Mar 27 '18 at 21:30
  • $\begingroup$ This said aluminum does form complexes with hydroxide, and aluminum metal reacts with base to form aluminate, $\ce{AlO2^2-}$ (and the hydroxide complexes). $\endgroup$ – ralk912 Mar 27 '18 at 21:33
  • $\begingroup$ The titration is not necessary to determine which hydroxide you have though! What other thing can you think of regarding these compounds (you already did the experiment for this part, actually). $\endgroup$ – ralk912 Mar 27 '18 at 21:38
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Your assumption of "the base reacted with the metal in aqueous solution" is not correct since you have filtered out metal from aqueous solution before you add $\ce {3M HCl}$. Therefore, I assume the error is from your calculations. Your answer, $\ce {13/1000 mol in 1.6 g}$ of beads, is off the mark. The correct answer should be $\ce {16.25 mmol of OH^- ions in 1.6 g}$. It means, either $\ce {\frac{16.25}{1} mmol of NaOH}$ or $\ce {\frac{16.25}{2} mmol of Ca(OH)_2}$ or $\ce {\frac{16.25}{3} mmol of Al(OH)_3}$ in $\ce {1.6 g}$ of beads. Therefore, your $\ce{\% NaOH}$ contained in beads is $\ce {\frac{16.25 mmol NaOH}{1} \cdot \frac{40 mg}{mmol of NaOH}\cdot \frac{1 g}{1000 mg}\cdot \frac{100 g of beads}{1.6 g of beads}}=40.6\%$ ($w/w$).

Similarly, your $\ce{\% Ca(OH)_2}$ contained in beads is $\ce {\frac{16.25 mmol Ca(OH)_2}{2} \cdot \frac{74 mg}{mmol of Ca(OH)_2}\cdot \frac{1 g}{1000 mg}\cdot \frac{100 g of beads}{1.6 g of beads}}=37.6\%$ ($w/w$), or your $\ce{\% Al(OH)_3}$ contained in beads is $\ce {\frac{16.25 mmol Al(OH)_3}{3} \cdot \frac{78 mg}{mmol of Al(OH)_3}\cdot \frac{1 g}{1000 mg}\cdot \frac{100 g of beads}{1.6 g of beads}}=26.4\%$ ($w/w$).

PS: Please also use unit $\ce{mol}$ to represent moles. The unit $\ce{M}$ is for $\ce{mol\cdot L^{-1}}$.

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  • $\begingroup$ Ok, thank you! I have recalculated the values. $\endgroup$ – JavaScriptCoder Mar 29 '18 at 12:56

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