Calculating the pH of 350L of Sulfuric Acid (H2SO4)

[chawki]

Homework Statement

Sulfuric acid (H₂SO₄) were collected from the waste acid container and the volume is 350 l.

Homework Equations

a) Calculate the acid pH of the solution when the hydrogen ion concentration is 2.0 ∙ 10^-4 mol / l.

The Attempt at a Solution

i think it's simple no ?
PH=-log H₃O⁺
PH= -log 2*10^-4
PH=3.69

 

[dacruick]

Maybe. Its either that simple or you have to take into account the auto ionization of water. Since your acid is at a low enough concentration, the water may come into play. Also, there is a chemistry forum that you might have more luck posting on.

EDIT: I can't remember for sure, but I think you're answer is right since its at 10^-7 mol/L that you have to take into account the water's pH

 

[chawki]

where did that 10^-7 came from? and why they gave us the volume of 350L

 

[dacruick]

I don't know. You need to find the concentration at which you need to take into account water's pH. I vaguely remember 10^-7 mol/L as being something important, but as I said, I can't remember for sure. Google it or something.

 

[Borek]

See pH of a strong acid/base.

The question is either that simple, or there is something more to it, a lot depends on what you were doing lately. For sure volume is irrelevant. It is possible that they wanted you to take ionic strength of the solution into account. If you have no idea what ionic strength and activity coefficients are (and you have not slept during a lecture), 3.96 is right (although it probably contains too many significant figures).

 

[chawki]

ok thank you, i think they ment only that 3.96
however in part B they asked :what is the mass of sodium hydroxide NaOH needed to neutralize the effect of the acid solution..?

 

[Borek]

Here the volume will come handy, as for stoichiometry you need number of moles.

 

[chawki]

Here the volume will come handy, as for stoichiometry you need number of moles.

is it ok like this:
2 NaOH + H₂SO₄ ----->Na₂SO₄ + 2 H₂O
m-----------350000g
2*40g------98.07g

m=285510.34g
that's 285.51 Kg

 

[Borek]

350 kg of solution is not equivalent to 350 kg of sulfuric acid.

 

[chawki]

350 kg of solution is not equivalent to 350 kg of sulfuric acid.

Oups

2 NaOH + H₂SO₄ ----> Na₂SO₄ + 2H₂O

m-------------------------350000g
2*40g--------------------178.07g

m=157241.53 g
that's 157.24 Kg

 

[Borek]

You are still completely wrong and guessing at random won't get you far.

350 kg is not mass of anything that you need to use. You need to use CONCENTRATION to calculate amount of substance.

 

[chawki]

I'm lost :(
can you please tell me the answer, and i will try to find out how it came..Please

 

[Borek]

No, I will try to walk you through the problem.

First of all - if the concentration of H⁺ is 2x10^-4 M, what is the concentration of sulfuric acid?

 

[chawki]

I really don't know but i can try this:
concentration of sulfuric acid = n/V

n= m/M = 350000/98.07 = 3568.87 moles

concentration of sulfuric acid = n/V = 3568.87/350 = 10.19 mol/l

 

[Borek]

You are again trying to calculate number of moles of acid assuming its mass is 350 kg - it is not. You have 350 kg of solution, not of the acid.

What is dissociation reaction of sulfuric acid in water? Can you write it? How many moles of H⁺ are produced for each mole of sulfuric acid put into water?

 

[chawki]

i don't know.

 

[Borek]

Seems like your teachers take money for nothing then.

Let's try different approach.

How many moles of NaOH are required to neutralize 1 mole of H⁺?

 

[chawki]

I really don't know, I'm sorry...
i have learned in french and this is kind of confusing for me..i hate to say this..but I'm going to give up about this problem..before it confuse the concepts that i already know and work with.

 

[Borek]

How does NaOH react with acid? What are products?

 

[chawki]

How does NaOH react with acid? What are products?

Na₂SO₄ and 2H₂O and H₃O⁺ i think

 

[Borek]

No, just salt and water. H₃O⁺ is a product of acid dissociation (H₃O⁺ or H⁺ - these are in many ways equivalent).

Do you know what net ionic reaction is?

Can you write balanced reaction equation for a reaction of H⁺ with NaOH?

 

[chawki]

Does this law help?
concentration of acid * volume of acid = concentration of base * volume of base

 

[Borek]

This is not a law and in general it is wrong - but it is not far from what you need to solve the question.

Take a look here: http://www.titrations.info/titration-calculation - while it is about titration, it nicely addresses everything needed to solve the problem.

 

[chawki]

i can't use much of Ca*Va = Cb*Vb
but i can write it: n_a = n_b
n_a = m_b/M_b
m_b = n_a*M_b
m_b = 2*40 = 80g

 

[Borek]

Stop using masses for everything.

Can you calculate number of moles of H⁺? You know concentration and volume.

 

[chawki]

n=C*V = 2*10^-4*350= 0.07mol

 

[Borek]

OK, now how many moles of NaOH will react with 0.07 moles of H⁺? Do you know what the reaction is? Reaction equation?

 

[chawki]

2 NaOH + H₂SO₄ ---> Na₂SO₄ + 2 H₂O

 

[Borek]

This reaction is correct, but it is not what I am asking about.

I am not asking about the reaction with sulfuric acid, but about reaction with H⁺.

 

[chawki]

Ahhh i don't know that one

 

[Borek]

Can you write NaOH in dissociated form?

 

[chawki]

sure
NaOH ----> Na⁺ + OH^-

 

[Borek]

OK, now, what would happen when you add H⁺ to the mix?

H⁺ + Na⁺ + OH^- -> ?

 

[chawki]

i don't know...but why we will add H⁺

 

[Borek]

Why? Because the question is about reaction between H⁺ and OH^-. You have a solution of H⁺ and you add solution of OH^- (Na⁺ and SO₄^2- don't take part in the reaction, they are so called spectators).

What simple, electrically neutral molecule can be a product of reaction between H⁺ and OH^-? Try to guess, it is so simple you will hate yourself you have not realized it earlier.