General Chemistry help needed

[golf10s]

Could anyone explain to me how to do this problem?

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

a) 10.2 mL
b) 20.2 mL
c) 30.4 mL
d) 35.5 mL
e) None of these

And the answer is e). It seemed like the answer was b) but I guess not [V1 = 1 x 10^-2 *(V1 + 20 mL)]

Thanks for the help.

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[MEDicated]

the answer is 20.4mL. you had the right equation, but after adding the additional amount into solution, the total volume is about 40 instead of 20.

 

[Balki]

This is the way I'd approach it:

In order to make a solution of pH of = 2 you need [H+] of 0.01 excess from whatever nutralization happened. This is because pH is ?log[H+]. You have a base and acid, both with normalities of one. You know you have 2 moles of the base and Hence you only need 3 moles of HCl so that the difference is one mole. 3 moles of HCl will be 30 ml of HCl. Molarity is one for both solutions.

So the answer is 30 ml. I certainly hope this helps, I wasn't quite familiar with the equation you had there.

 

[MEDicated]

ok, let me explain why it's 20.4 mL. First, you need to add enough HCl to neutralize the solution. b/c it is a 1M/1M ratio, adding 20 mL will give you a pH of 7. use the equation M1V1=M2V2. the desired molarity of the hydrogen ion is .01 b/c that will give a pH of 2. the concentration added is 1M (as stated in the question).

1M * (x mL) = (.01M) * (40 + x mL)

the reason there is 40 is b/c after neutralization, the solution has 40 mL (20 mL of base and 20 mL of acid). solve for x and then add that value to the 20 mL you already added to neutralize the solution. hope that clears it up. you should get 20.4mL

 

[Balki]

you are right, i didn't account for the increase in the volume.