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0.1 M HCL solution

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**clinicalMT**:

I'm a clinical lab dude, and it's been a log time (11 Years) since chem gen.

I'm trying to make a 0.1 M HCL solution to lyse RBC'S in body fluid. I came up with 1.0375ml HCL diluted to 125ml with H2O(volume needed). Is this correct?? If not please show your work so I can do it correctly in the furture.

**ssssss**:

I think your data is insufficient.I mean of what molarity of 1.035ml of HCL are you Diluting with 125ml of water.If you are having M1 molarity of 1.035ml of HCL and you are diluting it to V2 ml to obtain a Solution of M2 molarity.Then use this molarity equation:

M1xV1=M2xV2

M1x1.035=0.1xV2.

**movies**:

The concentrated HCl that you buy is about 12 M. If that's the stuff you are using then your calculation checks out. Here's my work:

12 M HCl * 0.0010375 L = 0.01245 mol HCl

0.01245 mol HCl / 0.125 L = 0.0996 M HCl

Probably close enough for your purposes. I think the conc. HCl is a little more than 12 M, but that's the number I was taught to remember. (BTW, if you assume the conc. HCl is 12.2 M then the solution would end up being 0.101 M. See this thread for why the 12.2M number matters: http://www.chemicalforums.com/index.php?board=2;action=display;threadid=857).

**clinicalMT**:

Thanks, that was a correct assumption, I was using 12 M HCL concentrated (diluting 1.0375 ml to 125 ml with H2O... sorry for not stating that in my original inquiry. Thanks again for checking my work and releaving my doubts!!

--- Quote from: movies on August 18, 2004, 12:38:04 PM ---The concentrated HCl that you buy is about 12 M. If that's the stuff you are using then your calculation checks out. Here's my work:

12 M HCl * 0.0010375 L = 0.01245 mol HCl

0.01245 mol HCl / 0.125 L = 0.0996 M HCl

Probably close enough for your purposes. I think the conc. HCl is a little more than 12 M, but that's the number I was taught to remember. (BTW, if you assume the conc. HCl is 12.2 M then the solution would end up being 0.101 M. See this thread for why the 12.2M number matters: http://www.chemicalforums.com/index.php?board=2;action=display;threadid=857).

--- End quote ---

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