Titration Problem

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

LoLPoPs

Full Member
10+ Year Member
7+ Year Member
What volume of HCl was added if 20mL of 1 M NaOH is titrated with 1 M HCl to produce a pH=2?

This problem is from TopScore Exam#2. I understand that 20mL of HCl will react with 20mL of NaOH. But where do I go from here? Can someone explain this in detail please? Thanks!!! drm85

Full Member
10+ Year Member
5+ Year Member
could you provide the answer chioces?

LoLPoPs

Full Member
10+ Year Member
7+ Year Member
could you provide the answer chioces?

Oooups sorry about that A) 10.2 mL

B) 20.2 mL

C) 30.4 mL

D) 35.5 mL

E) None of these

The answer is E) None of these...20.4mL.
But I'm not sure how to tackle these types of Q. alanan84

D1
10+ Year Member
5+ Year Member
Use M1V1=M2V2

M1=1M HCl, V1=x, M2=.01 (concentration of H+ at pH of 2), V2=40mL + x

(1M)(x)=(.01M)(40+x)
x=0.4mL

40.4 total mL - 20mL NaOH = 20.4mL HCl

alanan84

D1
10+ Year Member
5+ Year Member
Use M1V1=M2V2

M1=1M HCl, V1=x, M2=.01 (concentration of H+ at pH of 2), V2=40mL + x

(1M)(x)=(.01M)(40+x)
x=0.4mL

40.4 total mL - 20mL NaOH = 20.4mL HCl

I guess I could be a little clearer. You know that 20mL of HCl will react with 20mL NaOH to give a neutral pH. So with the equation you are finding x which is the volume of needed to bring it down to a pH of 2. Then you just add that to the amount of HCl added.

I need to go to sleep...

LoLPoPs

Full Member
10+ Year Member
7+ Year Member
I guess I could be a little clearer. You know that 20mL of HCl will react with 20mL NaOH to give a neutral pH. So with the equation you are finding x which is the volume of needed to bring it down to a pH of 2. Then you just add that to the amount of HCl added.

I need to go to sleep...

The more I think about it, I'm realizing this question requires some critical thinking. But it makes total sense now after reading your explanation. Thanks so much for your help! 