# Find the concentration of protons

• Lifeforbetter
In summary, if you have enough mathematics qualification, you can 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.
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?

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

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

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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?

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?

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?

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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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?

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?

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.

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?

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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.

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.

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

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

symbolipoint
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.

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?

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.

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

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+?

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?

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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##

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.

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.
x = 0.0665 M ?

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.

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

Lifeforbetter said:
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.

## 1. What is the definition of concentration of protons?

The concentration of protons refers to the number of protons present in a given volume or space. It is a measure of the amount of positive charge in a solution or substance.

## 2. Why is it important to find the concentration of protons?

Finding the concentration of protons is important in understanding the properties and behavior of substances. It can also help in determining the acidity or basicity of a solution, as well as the strength of an acid or base.

## 3. How is the concentration of protons measured?

The concentration of protons is typically measured using a pH scale, which ranges from 0 to 14. A pH of 7 is considered neutral, while lower values indicate acidity and higher values indicate basicity. The concentration of protons can also be measured using specialized equipment such as a pH meter or titration.

## 4. What factors can affect the concentration of protons?

The concentration of protons can be affected by a variety of factors, including the presence of other ions or molecules, temperature, and pressure. It can also be influenced by chemical reactions and the strength of acids and bases.

## 5. How is the concentration of protons related to the concentration of other ions?

The concentration of protons is closely related to the concentration of other ions in a solution. This is because protons can interact with other ions to form different compounds or affect the overall acidity or basicity of a solution. Changes in the concentration of protons can also lead to changes in the concentration of other ions.

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