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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nlong8 on May 06, 2009, 12:01:38 AM

Title: molar concentration problem, please help
Post by: nlong8 on May 06, 2009, 12:01:38 AM

Hello, rhis is a discussion problem in my lab,
if anyone could point me in the right direction it would be sublime.

Excess Silver NItrate was added to 14.04 mL of an unknown solution of HCL. The chloride was precipitated as AgCl. The AgCl was then dried and found to have a mass of .2758g. Calculate the Molar concentration of HCL
Thanks
Nate

Title: Re: molar concentration problem, please help
Post by: plankk on May 06, 2009, 12:06:32 AM

Start with the reaction equation.

Title: Re: molar concentration problem, please help
Post by: nlong8 on May 06, 2009, 12:27:25 AM

AgNo3 + HCl --> AgCl + H + N03
hows that?

Title: Re: molar concentration problem, please help
Post by: UG on May 06, 2009, 12:46:09 AM

Calculate the amount of moles of AgCl in the 0.2758g sample. Using n(moles)=mass(g)/Molar mass(g mol^-1)
The number of moles of AgCl equals the number of moles of HCl, look at your equation, it is a 1:1 ratio. rearrange this formula to find the concentration: n=c(mol L^-1) x V(L)

Title: Re: molar concentration problem, please help
Post by: nlong8 on May 06, 2009, 01:02:49 AM

.001924=.0104xmol/L
=.1374 M   
sound right?

Title: Re: molar concentration problem, please help
Post by: UG on May 06, 2009, 01:11:49 AM

Yep! Correct.  :D

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Quote from: nlong8 on May 06, 2009, 01:02:49 AM

.001924=.0104xmol/L
=.1374 M   
sound right?

Except it is 0.01404. I'm sure that was just a typo though.

Title: Re: molar concentration problem, please help
Post by: nlong8 on May 06, 2009, 01:18:53 AM

ah, you are amazing,
thank you
nate
 ;D