Gchem: Acid and Bases

[Envision]

Hi guys, ok so im having some trouble understanding titration problems. here are 2 examples.

1) What volume of HCl was added if 20 ml of 1M 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
answer: E
i noticed that the molarity of both are equal, so they should probably neutralize each other if given the same volume. however, im not sure where to go from there.

2) A 10 ml sample of HCl solution has a pH of 2. what volume of water must be added in order to change pH=4?

Answer: 990ml H2O

thanx in advance.

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[dat_student]

Hi guys, ok so im having some trouble understanding titration problems. here are 2 examples.

1) What volume of HCl was added if 20 ml of 1M 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
answer: E
i noticed that the molarity of both are equal, so they should probably neutralize each other if given the same volume. however, im not sure where to go from there.

2) A 10 ml sample of HCl solution has a pH of 2. what volume of water must be added in order to change pH=4?

Answer: 990ml H2O

thanx in advance.

Do a search on #1. I've probably solved it 10 times already

#2)
10^-2 * 10
______________ = 10^-4 (pH = 4 so [H+] = 10^-4)
10 + V

10^-1
_______ = 10 + V
10^-4

1000 = 10 + V

V = 990 ml


If you want to be more accurate you should consider that water itself has some H+. Since that amnt is small you can ignore it.

 

[DonExodus]

Do a search on #1. I've probably solved it 10 times already

#2)
10^-2 * 10
______________ = 10^-4 (pH = 4 so [H+] = 10^-4)
10 + V

10^-1
_______ = 10 + V
10^-4

1000 = 10 + V

V = 990 ml


If you want to be more accurate you should consider that water itself has some H+. Since that amnt is small you can ignore it.

I had always set it up like (.01)(.01)=(.0001)(V2) and solved from there, getting 100.

Ive seen you answer several questions in this format, but Im unclear how to follow it. I was just doing CV=CV, and I think this is a similar set up, but Im still lost.
Could you show me how you set those up please?
Thanks,
Don

 

[dat_student]

Do a search on #1. I've probably solved it 10 times already

#2)
10^-2 * 10
______________ = 10^-4 (pH = 4 so [H+] = 10^-4)
10 + V

10^-1
_______ = 10 + V
10^-4

1000 = 10 + V

V = 990 ml


If you want to be more accurate you should consider that water itself has some H+. Since that amnt is small you can ignore it.

I had always set it up like (.01)(.01)=(.0001)(V2) and solved from there, getting 100.

Ive seen you answer several questions in this format, but Im unclear how to follow it. I was just doing CV=CV, and I think this is a similar set up, but Im still lost.
Could you show me how you set those up please?
Thanks,
Don

# of moles of H+
______________ = molarity
total volume


# of moles of H+
_____________________________ = [H+] (i.e. concentration of H+)
volume of HCl + volume of water

 

[Envision]

Again, not to be stupid, but 1M HCL has a pH of 1, correct? pH is the [H+] = M, not the mol H+.
Whats confusing about this to me is that a 2M solution would have a pH of 2, which makes no sense, so I know my logic is flawed.

Hey Don,

1M HCl does not equal pH=1 and 2M HCl does not equal topH 2. 0.1 M HCl equals pH=1 and .001 M HCl equalys pH=2

1M HCl-> pH=-log(10^0)=0
0.1 M HCl -> pH=-log(10^-1)= 1
.002 M HCl -> pH=-log (10^-2)=2

Thanx DATstudent, i understand it now.

 

[joonkimdds]

2) A 10 ml sample of HCl solution has a pH of 2. what volume of water must be added in order to change pH=4?

Answer: 990ml H2O

thanx in advance.

let me solve it with my own way, i hope this makes it easier

pH = concentration of H
pH=2 = 10^(-2) = 0.1 = very very acidic
pH=4 = 10^(-4) = 0.001 = less acidic which means u added water here.

M1V1=M2V2
(0.1)(10mL) = (0.001)(10ml+x)
how?
becuz. 0.1 = molarity of H before u add water and so it's the 10ml HCL with pH=2.
and then it becomes pH=4 = 0.001 when u add water, x amount to 10mL.

and X is how much u added water and if u solve for it, u get 0.99mL and
that's 990L.

 

[gogiantsss]

Remember that in the lab, never add water to acid!

 

[joonkimdds]

Remember that in the lab, never add water to acid!

why not?

 

[Chga]

b/c its always acid to water

 

[Lonely Sol]

why not?

A large amount of heat is released when strong acids are mixed with water. Adding more acid releases more heat. If you add water to acid, you form an extremely concentrated solution of acid initially. So much heat is released that the solution may boil very violently, splashing concentrated acid out of the container! If you add acid to water, the solution that forms is very dilute and the small amount of heat released is not enough to vaporize and spatter it.

So Always Add Acid to water, and never the reverse, unless you want to get ur instructors attention!

 

[joonkimdds]

A large amount of heat is released when strong acids are mixed with water. Adding more acid releases more heat. If you add water to acid, you form an extremely concentrated solution of acid initially. So much heat is released that the solution may boil very violently, splashing concentrated acid out of the container! If you add acid to water, the solution that forms is very dilute and the small amount of heat released is not enough to vaporize and spatter it.

So Always Add Acid to water, and never the reverse, unless you want to get ur instructors attention!

I am so glad I made my table members do everything for me