How to Determine Proton Concentration in a Mixed Solution?

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To determine proton concentration in a mixed solution of H2SO4 and NaOH, the reaction stoichiometry indicates that one mole of NaOH reacts with one mole of H2SO4 to produce NaHSO4 and water. The formal concentration of HSO4- after the reaction is 0.5 M, but this must be adjusted for its dissociation as a weak acid, which has a pKa of 1.99. The equilibrium expression for the dissociation of HSO4- can be used to find the concentration of H3O+ in the solution. Understanding the difference between formality and molarity is crucial for accurate calculations in this context. Proper algebra and chemistry fundamentals are necessary to solve for the final concentrations accurately.
Lifeforbetter
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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?
 
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What effect does the reaction with NaOH have?
 
mjc123 said:
What effect does the reaction with NaOH have?
2NaOH + H2SO4 -> Na2SO4 + 2H2O
 
Do you have twice as much NaOH as H2SO4?
 
mjc123 said:
Do you have twice as much NaOH as H2SO4?
Mol of NaOH is 0.1 mol
I don't know about H2SO4
 
You have 100 mL of 1M. how many moles is that?
 
mjc123 said:
You have 100 mL of 1M. how many moles is that?
0.1 mol
 
Last edited:
mjc123 said:
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+ ?
 
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?
 
  • #10
mjc123 said:
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?
 
  • #11
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?
 
  • #12
mjc123 said:
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?
 
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  • #13
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?
 
  • #14
mjc123 said:
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?
 
  • #15
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 Ka 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.
 
  • #16
symbolipoint said:
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 Ka 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?
 
Last edited:
  • #17
Lifeforbetter said:
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 pKa.
The acid dissociation constant for bisulfate ion is 1.02x10-2.
 
  • #18
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.
 
  • #19
I don't understand.. so in order to count proton concentration. What its required?
 
  • #20
No. I have to understand the basic theory first. I will come back.
 
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Likes symbolipoint
  • #21
Lifeforbetter said:
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.
 
  • #22
symbolipoint said:
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 Ka 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?
 
  • #23
Lifeforbetter said:
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 said:
... 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.
 
  • #24
Ok.
H2SO4 + NaOH -> NaHSO4 + H2O
H = 0.1 mol / 0.2 L = 0.5 M
 
  • #25
No. The formal concentration of HSO4- is 0.5M. You have to consider the dissociation
HSO4- + H2O → H3O+ + SO42-
You have been given the equilibrium constant for this reaction. Can you use it to calculate the concentration of H3O+?
 
  • #26
mjc123 said:
No. The formal concentration of HSO4- is 0.5M. You have to consider the dissociation
HSO4- + H2O → H3O+ + SO42-
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+ + SO42- ?
What is concentration of H2O?
 
Last edited:
  • #27
symbolipoint said:
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##
 
  • #28
x is the concentration of H3O+, which is equal to the concentration of SO42-. 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.
 
  • #29
mjc123 said:
x is the concentration of H3O+, which is equal to the concentration of SO42-. 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 ?
 
  • #30
mjc123 said:
No. The formal concentration of HSO4- is 0.5M. You have to consider the dissociation
HSO4- + H2O → H3O+ + SO42-
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.
 

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