Find the concentration of protons

[Lifeforbetter]

Homework Statement:

NaOH 1M 100 mL + H2SO4 1M 100mL
Find the concentration of proton after mixed?

Relevant Equations:

H2SO4 + H2O -> H3O+ + HSO4-

After mixed, the volume is 200mL
I am confused. The mol of H2SO4 is M*V
1M*100mL or 1M*200mL ?
Finding the concentration of proton means the Molarity for H3O+ right? mol/volume
H2SO4 + H2O -> H3O+ + HSO4-
Mol H3O+ = Mol H2SO4?

 

[mjc123]

What effect does the reaction with NaOH have?

 

[Lifeforbetter]

What effect does the reaction with NaOH have?

2NaOH + H2SO4 -> Na2SO4 + 2H2O

 

[mjc123]

Do you have twice as much NaOH as H₂SO₄?

 

[Lifeforbetter]

Do you have twice as much NaOH as H₂SO₄?

Mol of NaOH is 0.1 mol
I don't know about H2SO4

 

[mjc123]

You have 100 mL of 1M. how many moles is that?

 

[Lifeforbetter]

You have 100 mL of 1M. how many moles is that?

0.1 mol

 

[Lifeforbetter]

Do you have twice as much NaOH as H₂SO₄?

Do you mean i need NaOH twice as much as H2SO4? So there's 0.5 mol excess of H2SO4? And produces H3O+ ?

 

[mjc123]

The reaction you wrote in #3 requires twice as much NaOH as H₂SO₄. What reaction occurs if you have the same amount of NaOH as H₂SO₄?

 

[Lifeforbetter]

The reaction you wrote in #3 requires twice as much NaOH as H₂SO₄. What reaction occurs if you have the same amount of NaOH as H₂SO₄?

Some H2SO4 will be left. All NaOH reacted?

 

[mjc123]

Which do you think is more likely - half the H₂SO₄ reacts completely, or all of it reacts half-way? Can you write an equation for the equimolar reaction of NaOH and H₂SO₄?

 

[Lifeforbetter]

Which do you think is more likely - half the H₂SO₄ reacts completely, or all of it reacts half-way? Can you write an equation for the equimolar reaction of NaOH and H₂SO₄?

2NaOH + H2SO4 -> Na2SO4 + 2H2O
0.1 mol---0.1mol-----0mol--------0mol
x mol-----0.5xmol----0.5xmol----0.5xmol
(0.1 - x)mol 0.1-(0.5x)mol

If it reacts halfway then
H2SO4 + NaOH -> NaHSO4 + H2O
H+ + H2O -> H3O+
Is it?

 

[mjc123]

Yes, that is correct. HSO₄^- is a weak acid with a pKa of 1.99. Can you work out the proton concentration in the solution?

 

[Lifeforbetter]

Yes, that is correct. HSO₄^- is a weak acid with a pKa of 1.99. Can you work out the proton concentration in the solution?

Is it 0.5 mol?

 

[symbolipoint]

Have you figured yet that resulting "moles" of HSO4^-1 becomes 'first' 0.1 moles, and that Ka=10^^(-1.99) ? One mole of NaOH reacts with one mole H₂SO₄ to give one mole of the bisulfate ion. The K_a is weak but not too weak. You could try an equilibrium calculation if you want.

You may continue with formal concentration of 0.1moles bisulfate per 0.2 liters solution; what is that in formality?

If x is the molarity of hydronium ion then you can form (x^²)/(0.5-x)=10^^-1.99 and solve the simple quadratic equation.

 

[Lifeforbetter]

Have you figured yet that resulting "moles" of HSO4^-1 becomes 'first' 0.1 moles, and that Ka=10^^(-1.99) ? One mole of NaOH reacts with one mole H₂SO₄ to give one mole of the bisulfate ion. The K_a is weak but not too weak. You could try an equilibrium calculation if you want.

You may continue with formal concentration of 0.1moles bisulfate per 0.2 liters solution; what is that in formality?

If x is the molarity of hydronium ion then you can form (x^²)/(0.5-x)=10^^-1.99 and solve the simple quadratic equation.

I don't understand your first question..

concentration of HSO4 = 0.5 M ?
Do you mean that the concentration of HSO4 that left after reacted with NaOH?

Concentration of proton = concentration of H3O⁺?

If x is the molarity of hydronium ion then you can form (x^²)/(0.5-x)=10^^-1.99 how to get to this equation?

 

[symbolipoint]

I don't understand your first question..

concentration of HSO4 = 0.5 M ?
Do you mean that the concentration of HSO4 that left after reacted with NaOH?

Concentration of proton = concentration of H3O⁺?

If x is the molarity of hydronium ion then you can form (x^²)/(0.5-x)=10^^-1.99 how to get to this equation?

My first question: Yes, but it does not stay that way. HSO4^-1 is a weak acid (not too weak) and it dissociates. The formal concentration is 0.5 F but the MOLAR concentration becomes less because the acid dissociates.

Look again at that pK_a.
The acid dissociation constant for bisulfate ion is 1.02x10^-2.

 

[symbolipoint]

Let me write the quadratic equation a little better:
(x²)/(0.5-x)=1.02*10^-2

The asterisk is the multiplication symbol to avoid confusing the "x" symbols.

 

[Lifeforbetter]

I don't understand.. so in order to count proton concentration. What its required?

 

[Lifeforbetter]

No. I have to understand the basic theory first. I will come back.

 

[symbolipoint]

No. I have to understand the basic theory first. I will come back.

Good Idea. If enrollment in a community college is your option, you could, if you have enough Mathematics qualification, study "Elementary Chemistry" to build some of the basics of Stoichiometry and understanding of some simpler inorganic reaction knowledge. Then again if you have the additional Math requirements, you can continue on to study General Chemistry; and in either first or second of the semesters of it, you will study equilibrium of weak acids and bases.

 

[Lifeforbetter]

Have you figured yet that resulting "moles" of HSO4^-1 becomes 'first' 0.1 moles, and that Ka=10^^(-1.99) ? One mole of NaOH reacts with one mole H₂SO₄ to give one mole of the bisulfate ion. The K_a is weak but not too weak. You could try an equilibrium calculation if you want.

You may continue with formal concentration of 0.1moles NaHSO₄ per 0.2 liters solution; what is that in formality?

If x is the molarity of hydronium ion then you can form (x^²)/(0.5-x)=10^^-1.99 and solve the simple quadratic equation.

Ok. So 0.1 mol of NaOH and 0.1 mol of H₂SO₄ reacted half way.. to give 0.1 mol of NaHSO₄.
So what reaction happen next? Since there is no NaOH left. My goal is to find concentration of H⁺ in NaHSO₄?

 

[symbolipoint]

Ok. So 0.1 mol of NaOH and 0.1 mol of H₂SO₄ reacted half way.. to give 0.1 mol of NaHSO₄.
So what reaction happen next? Since there is no NaOH left. My goal is to find concentration of H⁺ in NaHSO₄?

... If enrollment in a community college is your option, you could, if you have enough Mathematics qualification, study "Elementary Chemistry" to build some of the basics of Stoichiometry and understanding of some simpler inorganic reaction knowledge. Then again if you have the additional Math requirements, you can continue on to study General Chemistry; and in either first or second of the semesters of it, you will study equilibrium of weak acids and bases.

 

[Lifeforbetter]

Ok.
H2SO4 + NaOH -> NaHSO4 + H2O
H = 0.1 mol / 0.2 L = 0.5 M

 

[mjc123]

No. The formal concentration of HSO₄^- is 0.5M. You have to consider the dissociation
HSO₄^- + H₂O → H₃O⁺ + SO₄^2-
You have been given the equilibrium constant for this reaction. Can you use it to calculate the concentration of H₃O⁺?

 

[Lifeforbetter]

No. The formal concentration of HSO₄^- is 0.5M. You have to consider the dissociation
HSO₄^- + H₂O → H₃O⁺ + SO₄^2-
You have been given the equilibrium constant for this reaction. Can you use it to calculate the concentration of H₃O⁺?

The coefficient between HSO₄^- and H₃O⁺ are the same? (The concentration not the same? That is 0.5 M?) What is F? Isn't mole/volume = M ?
Which equation to use equilibrium constant formula to find concentration of H₃O⁺ from this equation? HSO₄^- + H₂O → H₃O⁺ + SO₄^2- ?
What is concentration of H2O?

 

[Lifeforbetter]

Let me write the quadratic equation a little better:
(x²)/(0.5-x)=1.02*10^-2

I don't know where this come from.. 0.5 - x and x refers to what concentration..
But i calculate
x = 0.01 +- 0.05$\sqrt2$

 

[mjc123]

x is the concentration of H₃O⁺, which is equal to the concentration of SO₄^2-. 0.5-x is the concentration of HSO₄^-. We ignore the concentration of water as it is effectively constant.
Your solution to the quadratic equation is incorrect.

 

[Lifeforbetter]

x is the concentration of H₃O⁺, which is equal to the concentration of SO₄^2-. 0.5-x is the concentration of HSO₄^-. We ignore the concentration of water as it is effectively constant.
Your solution to the quadratic equation is incorrect.

x = 0.0665 M ?

 

[symbolipoint]

No. The formal concentration of HSO₄^- is 0.5M. You have to consider the dissociation
HSO₄^- + H₂O → H₃O⁺ + SO₄^2-
You have been given the equilibrium constant for this reaction. Can you use it to calculate the concentration of H₃O⁺?

Just in being fair, he may need to reread or recheck the distinction between FORMALITY and MOLARITY.

 

[symbolipoint]

x is the concentration of H₃O⁺, which is equal to the concentration of SO₄^2-. 0.5-x is the concentration of HSO₄^-. We ignore the concentration of water as it is effectively constant.
Your solution to the quadratic equation is incorrect.

(I did not solve it. I only wrote it.)

 

[symbolipoint]

I don't know where this come from.. 0.5 - x and x refers to what concentration..
But i calculate
x = 0.01 +- 0.05$\sqrt2$

Study basic algebra: Algebra 1 and Algebra 2;
then study Elementary Chemistry, and the first or the first AND second semesters of General Chemistry; only then may you understand these topics.