acid/base chemistry question

[dentalstudent2021]

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

so i know m1v1 = m2v2 would not work since its not the same substance being diluted.
I am having a hard time understanding the approach.

 

[510586]

Can you confirm if the answer is 30 mL? I just did it but not sure if I did it correctly.
I did pH=2 so that means H+ = .01 mol.
Then 20 mL * 1 mol/L NaOH gives you .02 mol OH-, as it dissociates completely b/c strong base. So .02 mol H+ would neutralize it, so you need another .01 mol H+ (.03 mol HCl dissociating completely) to get that pH=2? Is this correct?

 

[dentalstudent2021]

Can you confirm if the answer is 30 mL? I just did it but not sure if I did it correctly.
I did pH=2 so that means H+ = .01 mol.
Then 20 mL * 1 mol/L NaOH gives you .02 mol OH-, as it dissociates completely b/c strong base. So .02 mol H+ would neutralize it, so you need another .01 mol H+ (.03 mol HCl dissociating completely) to get that pH=2? Is this correct?

no, its 20.4 ml

 

[510586]

oh well I was way off my bad.

 

[510586]

Did you or anyone else find the way to do this question?

 

[cc609]

So this is how I did it and I got 20.4. If there is an easier way let us know!

There's a strong acid and a strong base with the same molarity. So basically the first part of the above post is right, you need to add 20mL to get it to be neutral. SO the TOTAL volume is 40 mL now. Now since it is acidic you know you need to add additional HCls

So pH= 2 will be .01M and x will be the amount you add in IN ADDITION to the 20mL for neutralisation, and 40+x will be the total volume.
I solved it algebraically w M1V1=M2V2

(1M)(x mL) = (.01M)(40+x)
x = 0.4 + 0.01x
.99x = 0.4
x = .4 mL
And since total all together is 40.4, subtract the original -20 mL of the base and you get 20.4 for the acid you added in. I hope that helps a little bit..