titration/dilution problem

[dental17]

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Problem:

If 10 mL of 1 M NaOH is titrated with 1 M HCl to a pH of 2, what volume of HCl was added.

Solution:

Since both acid and base are 1 M, 10 mL of HCl will neutralize 10 mL of NaOH. This produces 20 mL of 0.5 M NaCl solution.

1M HCl x X mL = 0.01 M x (20 + X) mL

X = additional HCl being added.
0.01 M = from the pH value of 2.

So X = 0.2 mL, adding that to the 10 mL gives 10.2 mL of HCl.

I understand where the 20 mL is coming from and the following calculation to get the final answer of 10.2 mL of HCL, but I don't understand why the solution is 0.5M.

 

[rockclock]

Problem:

If 10 mL of 1 M NaOH is titrated with 1 M HCl to a pH of 2, what volume of HCl was added.

Solution:

Since both acid and base are 1 M, 10 mL of HCl will neutralize 10 mL of NaOH. This produces 20 mL of 0.5 M NaCl solution.

1M HCl x X mL = 0.01 M x (20 + X) mL

X = additional HCl being added.
0.01 M = from the pH value of 2.

So X = 0.2 mL, adding that to the 10 mL gives 10.2 mL of HCl.

I understand where the 20 mL is coming from and the following calculation to get the final answer of 10.2 mL of HCL, but I don't understand why the solution is 0.5M.

10 mL of 1 M HCl yields 0.010 mol Cl-
10 mL of 1 M NaOH yields 0.010 mol Na+
Together, that's 0.010 mol NaCl in 20 mL = 0.5 M NaCl

 

[dental17]

I still don't get it 🙁 How are you getting 0.010 mol of Na+ and Cl-? Also, since the 0.5M never shows up in the equation is it even important in solving the problem?

10 mL of 1 M HCl yields 0.010 mol Cl-
10 mL of 1 M NaOH yields 0.010 mol Na+
Together, that's 0.010 mol NaCl in 20 mL = 0.5 M NaCl

 

[JustwantDDS]

M= mol/L so u have mL and M, solve for mol.
butttt i dont understand the 20 mL
and how you set up

1M HCl x X mL = 0.01 M x (20 + X) mL

 

[dental17]

Thanks for the mol explanation...but I still can't see why the solution is 0.5M

As for the 20 mL, they're coming from the original question that indicates 10 mL of HCl was added to neutralize 10 mL of NaOH giving us a total of 20 mL.

The equation follows the general formula M1V1=M2V2
(1 M HCl)(x mL) = (.01 M) [(20 + x) mL)
x= amount of HCl added
20 + x= total volume of solution

M= mol/L so u have mL and M, solve for mol.
butttt i dont understand the 20 mL
and how you set up

1M HCl x X mL = 0.01 M x (20 + X) mL

 

[rockclock]

I still don't get it 🙁 How are you getting 0.010 mol of Na+ and Cl-? Also, since the 0.5M never shows up in the equation is it even important in solving the problem?

No, it's not really relevant for this problem, but you should be able to see how you get 0.5 M NaCl because the concept itself is important for other dilution problems.

Molarity (M) is mol/L, so 1 M solution is 1 mol/1 L solution...

10 mL x (1 L/1000 mL) x (1 mol/1 L) = 0.010 mol

It's the same calculation for both Na+ and Cl- since both come from 10 mL of 1 M solution. 0.010 mol Na+ and 0.010 mol Cl- together is 0.010 mol NaCl. Your total volume is 20 mL.

0.010 mol/20 mL x (1000 mL/1 L) = 0.5 mol/L = 0.5 M.

Or you can view it as taking 10 mL of "1 M NaCl" (from a 1 M Na+ solution and a different 1 M Cl- solution) and diluting it to 20 mL total. Doubling the volume would halve the molarity, so 0.5 M.

I don't mean this offensively, but this really is very basic. Make sure you review this until you're solid otherwise it will, for sure, come back to screw you.

 

[dental17]

Thanks, rockclock. Btw, no offense taken--I haven't seen these kinds of problems in a really long time so even the basics are not so basic to me. Will definitely review until they solidify in my brain 🙂. Thanks for all your help!

No, it's not really relevant for this problem, but you should be able to see how you get 0.5 M NaCl because the concept itself is important for other dilution problems.

Molarity (M) is mol/L, so 1 M solution is 1 mol/1 L solution...

10 mL x (1 L/1000 mL) x (1 mol/1 L) = 0.010 mol

It's the same calculation for both Na+ and Cl- since both come from 10 mL of 1 M solution. 0.010 mol Na+ and 0.010 mol together is 0.010 mol NaCl.

0.010 mol/20 mL x (1000 mL/1 L) = 0.5 mol/L = 0.5 M.

Or you can view it as taking 10 mL of 1 M NaCl and diluting it to 20 mL. Doubling the volume would halve the molarity, so 0.5 M.

I don't mean this offensively, but this really is very basic. Make sure you review this until you're solid otherwise it will, for sure, come back to screw you.

 

[rockclock]

Thanks, rockclock. Btw, no offense taken--I haven't seen these kinds of problems in a really long time so even the basics are not so basic to me. Will definitely review until they solidify in my brain 🙂. Thanks for all your help!

No problem, happy to help...keep 'em coming. Good luck!