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 January 22nd, 2008, 07:49 PM #1 Member   Joined: Oct 2007 From: CA, USA Posts: 46 Thanks: 0 Chemistry Problem A student added 50.0 ml of NaOH solution to 100.0 ml of .400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in the formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution. Apparently, the answer is 2.00 M, but I don't get why. I can't figure out what the precipitate is and thus can't balance the equation. NaOH + HCl --> NaCl + H2O NaCl + H2O + Cr(NO3)3 --> ? If you have some clue about how to do this problem, please post! This is a homework problem and if anyone has any idea how to figure this out, then I'll probably be able to figure it out from there.
 June 4th, 2008, 10:41 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,289 Thanks: 1681 I'm not a chemist, but suspect the reaction is 3NaOH + Cr(NO3)3 --> 3NaNO3 + Cr(OH)3, and the chromium hydroxide was dried before being weighed.
 March 23rd, 2009, 10:20 PM #3 Member   Joined: Mar 2009 From: Asheville, NC Posts: 48 Thanks: 0 Re: Chemistry Problem Solve for what's soluble after the reaction... $NO_{3}$, nitrates, are always soluble, therefore, that's not the precipitate. It's too late at night to figure this out, so have a shot at it using your solubility rules and eliminate everything that hasn't changed state from one side to the other of the reaction... Also, writing $H_{2}O$ as $HOH$ helps sometimes. However, $3NaOH + Cr(NO_{3})_{3} -->{} 3NaNO_{3} + Cr(OH)_{3}$ sounds right. Simple double-displacement reaction... The nitrate was aqueous prior to the reaction and after the reaction, so you're left with the ionic equation $3OH^{-} + Cr^{3+} --> Cr(OH)_{3}$ So, 2.06g of $Cr(OH)_{3}$ precipitate, at a molar mass of 52 + 16*3 + 1*3 = 103g/mol. so, you have $2.06g/103g/mol= .02mol$ of $Cr(OH)_{3}$. based on this equation: $3OH^{-} + Cr^{3+} --> Cr(OH)_{3}$, you have 3mol of $OH^{-}$ reacting with 1mol of $Cr^{3+}$, with the result of 1mol of $Cr(OH)_{3}$. Substitute your values with .02mol. gives you .06mol total of OH^{-} Now, you have the equation $NaOH + HCl --> NaCl + HOH$ Now you have to solve for the amount of mole relation you have here... 100mL of .400M HCl = .04mol of HCl Remember that we have a total of .06mol of OH, based on the actual amounts of Cr(OH)_{3} created. And it is at this point that I am confused, it is 2:20 in the morning.... and I realize topic is from over a year ago. I'll go kill myself now *sigh*
April 10th, 2009, 05:21 PM   #4
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Re: Chemistry Problem

Quote:
 Originally Posted by mahuirong A student added 50.0 ml of NaOH solution to 100.0 ml of .400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in the formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution. Apparently, the answer is 2.00 M, but I don't get why. I can't figure out what the precipitate is and thus can't balance the equation. NaOH + HCl --> NaCl + H2O NaCl + H2O + Cr(NO3)3 --> ? If you have some clue about how to do this problem, please post! This is a homework problem and if anyone has any idea how to figure this out, then I'll probably be able to figure it out from there.
It would seem the chemist forgot to mention that most of theirs conception of reality is base upons not only theirs experiments but also a mixture of imaginations to which its realiability with reality seem in-so-far fit.

 April 11th, 2009, 06:57 AM #5 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: Chemistry Problem Vu-- that is an unhelpful comment. Please refrain from epistemological discussions on threads that are dedicated to other topics.

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# a student added 50.0 ml of an naoh solution to 100 ml of .400 m hcl and it created a percipate of 2.06g. what is the concentration of naoh

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