Topscore Gchem question.

[kponenation]

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What volume of HCl was added if 20mL of 1M of NaOH is titrated with 1M of HCl to produce pH=2?

I get that you need 20mL of HCl to neutralize the solution meaning pH of 7.
From there how do you solve the problem?

Thanks in advance.

 

[nartnad]

So you need 20 mL of the HCl to make it neutral. Keep that in mind, so you can add it to the value of "x", which is the amount of EXTRA HCl that is needed to make the pH lower to 2.

So in a neutral solution, we would have 20mL NaOH + 20mL HCl = 40mL
The volume of the final solution (of pH=2), therefore = .040L + x

Now we can use MV=MV, where:
(molarity final solution)(volume final solution)=(molarity original HCl)(x)
(1x10^-2 M)(.04L + x)=(1 M)x
x= .0004L, where x is the amount of extra HCl that is needed to lower the pH

So going back to the original problem, we needed to add 20mL HCl to make the solution neutral. Now we add "x" to make it acidic. So .02L + .0004L = .00204L

which is 20.4mL

Hope that helps

 

[kponenation]

Thanks for your explanation!

 

[fsddsms]

What volume of HCl was added if 20mL of 1M of NaOH is titrated with 1M of HCl to produce pH=2?

I get that you need 20mL of HCl to neutralize the solution meaning pH of 7.
From there how do you solve the problem?

Thanks in advance.

Ok so you got 20 mL HCl to get it to pH = 7. So there's 20 mL HCl/ 20 mL NaOH.
Next you set up another M1V1 = M2V2:
(1M HCl)(X mL) = (10^-2)( 40+ X mL)
X = 0.4 ml = amount added on top of the first 20 mL acid you already added to neutralize.
Note: There are 40.4 mL of liquid (acid with base) total in solution.
So 20.4mL HCl of acid to get it to pH of 2.

A good equation to know here is [H+] = 10^-ph --> [H+] at a pH of 2 = 10^-2 which is what we plug in for M2.

Hope that helps!!