Volumetric Analysis

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Miriam Villarreal 1J
Posts: 105
Joined: Sat Aug 17, 2019 12:16 am

Volumetric Analysis

Postby Miriam Villarreal 1J » Sun Oct 06, 2019 5:44 pm

L37) Part b: How many milliliters of 0.20M NaOH(aq) could be neutralized by 100.mL of the diluted solution?

Jamie Lee 1H
Posts: 54
Joined: Fri Aug 09, 2019 12:15 am

Re: Volumetric Analysis

Postby Jamie Lee 1H » Sun Oct 06, 2019 5:57 pm

You would need to use the equation M1V1= M2V2 for a problem that involves dilution. So for this question, you use the 0.50 M given in part A with and the 100ml converted to 1 L and solve for Vol NaOH
((1.00 L)(0.50M))/0.20 M= 2.5 x 10^2 mL

Tauhid Islam- 1H
Posts: 64
Joined: Fri Aug 02, 2019 12:15 am

Re: Volumetric Analysis

Postby Tauhid Islam- 1H » Sun Oct 06, 2019 6:09 pm

I think the 100. mL of the diluted solution is referring to 1.00 L of 0.50 m HNO 3 (aq) from part a. So if you have 1L of 0.50M of HNO3, and you are taking 100mL or .1L of that solution and you are trying to figure out how many mL of NaOH you need to neutralize the solution. If you just have HNO3 in solution, the molecule would disassociate into H+ and NO3-, making the solution acidic. Adding the NaOH in the solution would neutralize the acid because the NaOH would disassociate into Na+ and OH-, so the H+ and the OH- would react together to form H2O and that neutralizes the solution. Because for every 1 mole of HNO3, you get 1 mole of H+ and for every 1 mole NaOH, you get 1 mole of OH-, the moles of NaOH and HNO3 have to match up. So use the equation M1V1=M2V2 and solve for V2 to figure out the ml of 0.20M of NaOH(aq) you need. M1'= 0.50M of HNO3 and V1= .1L. I'm not entirely sure if this is correct.

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