Topscore 2 #68

[JNew]

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Could somebody please explain the solution to this problem:
What volume fo HCl was added if 20ml of 1M NaOH is titrated with 1M HCl to produce a pH=2?

The answer given is 20.4 ml.

Thanks

 

[UndergradGuy7]

Could somebody please explain the solution to this problem:
What volume fo HCl was added if 20ml of 1M NaOH is titrated with 1M HCl to produce a pH=2?

The answer given is 20.4 ml.

Thanks

20 mL of 1 M NaOH needs 20 mL of 1 M HCl since both are the same concentration.

So we need 20 mL to neutralize and then some more to make it pH of 2.

pH of 2 = 10^-2 = concentration of H+

so (1M HCl)(x mL)=(10^-2)(40+x)
I put 40+x because this is the total volume of the solution 20 mL from HCl + 20 mL from NaOH + x more we need to get it to pH of 2.

Solve for x you get 0.404 mL

so total mL of HCl needed is 20+0.404 mL = 20.404 mL

 

[herkulease]

this problem sure comes up a lot.

 

[JNew]

Still don't really understand why you set it up like this:
(1M HCl)(x mL)=(10^-2)(40+x)

Any way you could maybe explain in more detail?

 

[UndergradGuy7]

Use m1v1=m2v2 equation I am sure you have seen before.

m=molarity
v=volume

(1M HCl)(x mL)=(10^-2)(40+x)

So we need to add HCl from a container of HCl to our solution titration to get a pH of 2.

So left hand side (1M HCl)(x mL) is the container of HCl. It is 1 M and we don't know how much volume we need from it so x mL.

Right hand side
(10^-2)(40+x)
We know we want a concentration of 10^-2 M of H+
(40+x) is the total volume in the solution.
The volume is the tricky part because when you are adding HCl you keep changing the volume hence the +x. The 40 is from 20 mL HCl you added to the 20 mL NaOH to neutralize it. The +x is the mL you need to get to get pH 2.

 

[JNew]

Ohhhhhhhh, totally get it now. Thanks for the great explanation!
Is that a normal DAT type question btw?

Use m1v1=m2v2 equation I am sure you have seen before.

m=molarity
v=volume

(1M HCl)(x mL)=(10^-2)(40+x)

So we need to add HCl from a container of HCl to our solution titration to get a pH of 2.

So left hand side (1M HCl)(x mL) is the container of HCl. It is 1 M and we don't know how much volume we need from it so x mL.

Right hand side
(10^-2)(40+x)
We know we want a concentration of 10^-2 M of H+
(40+x) is the total volume in the solution.
The volume is the tricky part because when you are adding HCl you keep changing the volume hence the +x. The 40 is from 20 mL HCl you added to the 20 mL NaOH to neutralize it. The +x is the mL you need to get to get pH 2.