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Calculating the Hydrogen Ion Concentration of Strong Monoprotic Acids
pH is measure of the hydrogen ion concentration in a solution:
pH = log_{10}[H^{+}_{(aq)}]
We can rearrange this equation to find the concentration of hydrogen ions in a solution given its pH:
[H^{+}] = 10^{pH}
There are 2 steps for calculating the concentration of hydrogen ions (or hydronium ions H_{3}O^{+}) in a solution if you have been given the pH of the solution:
Step 1. Write the equation for finding [H^{+}]:
[H^{+}] = 10^{pH}
Step 2. Substitute in the value for pH and solve to give the concentration of H^{+} in mol L^{1}
Calculating the Concentration of Strong Monoprotic Acids
If we know the pH of a solution of strong acid, we can use this to calculate the concentration of the acid.
Step 1. Write the equation for finding [H^{+}]:
[H^{+}] = 10^{pH}
Step 2. Substitute in the value for value for pH and solve to give the concentration of H^{+} in mol L^{1}
Step 3. Write the equation for the complete dissociation of the strong monoprotic acid:
 monoprotic acid  →  hydrogen ions  +  anion 
hydrochloric acid  HCl  →  H^{+}  +  Cl^{} 
hydrobromic acid  HBr  →  H^{+}  +  Br^{} 
hydroiodic acid  HI  →  H^{+}  +  I^{} 
nitric acid  HNO_{3}  →  H^{+}  +  NO_{3}^{} 
perchloric acid  HClO_{4}  →  H^{+}  +  ClO_{4}^{} 

Strong Monoprotic Acids 
hyrochloric acid  HCl 
hydrobromic acid  HBr 
hyroiodic acid  HI 
nitric acid  HNO_{3} 
perchloric acid  HClO_{4} 

Step 4. Use the concentration of the hydrogen ions in solution to determine the concentration of the acid :
For a monoprotic acid, the stoichiometric ratio (mole ratio) of the acid, HA, to the hydrogen ions, H^{+}, is 1 : 1
general monoprotic acid :  HA  →  H^{+}  +  A^{} 
for 1 mole of acid :  1 mole HA  →  1 mole H^{+}  +  1 mole A^{} 
for 0.1 mole of acid :  0.1 mole HA  →  0.1 mole H^{+}  +  0.1 mole A^{} 
for 0.5 mole of acid :  0.5 mole HA  →  0.5 mole H^{+}  +  0.5 mole A^{} 
for 2.3 mole of acid :  2.3 mole HA  →  2.3 mole H^{+}  +  2.3 mole A^{} 
so for n mole of acid :  n mole HA  →  n mole H^{+}  +  n mole A^{} 
Concentration in mol L^{1} (molarity or molar concentration) is calculated by dividing moles by volume in litres:
molarity = moles ÷ volume
The volume of the solution is the same for both the undissociated acid, HA, and for the hydrogen ions, H^{+}, it produces.
general monoprotic acid :  HA  →  H^{+}  +  A^{} 
for n mole of acid in 1 L of solution:  [HA]=n/1  →  [H^{+}]=n/1  +  [A^{}]=n/1 
for n mole of acid in 2 L of solution:  [HA]=n/2  →  [H^{+}]=n/2  +  [A^{}]=n/2 
for n mole of acid in 0.4 L of solution:  [HA]=n/0.4  →  [H^{+}]=n/0.4  +  [A^{}]=n/0.4 
for n mole of acid in 1.3 L of solution:  [HA]=n/1.3  →  [H^{+}]=n/1.3  +  [A^{}]=n/1.3 
so for n mole of acid in V L of solution:  [HA]=n/V  →  [H^{+}]=n/V  +  [A^{}]=n/V 
We can see that the concentration of the hydrogen ions produced by the strong monoprotic acid will be the same as the concentration of the acid.
[HA] = [H^{+}] = 10^{pH}
Worked Examples
(based on the StoPGoPS approach to problem solving in chemistry.)
Question 1. Find the concentration of hydrogen ions in an aqueous solution of hydrochloric acid with a pH of 2.0
 What have you been asked to do?
Calculate the concentration of hydrogen ions
[H^{+}] = ? mol L^{1}
 What information (data) have you been given?
Extract the data from the question:
pH = 2.0
 What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding [H^{+}]:
[H^{+}] = 10^{pH}
 Substitute in the value for pH and solve:
[H^{+}] = 10^{pH}
[H^{+}] = 10^{2.0}
= 0.010 mol L^{1}
 Is your answer plausible?
Use your calculated value for [H^{+}] to find pH and compare it to that given in the question:
pH = log_{10}[H^{+}] = log_{10}[H^{+}] = 2
Since this value is the same as that given in the question we are confident our answer is correct.
 State your solution to the problem:
[H^{+}] = 0.010 mol L^{1}
Question 2. An aqueous solution of hydrochloric acid has a pH of 3.6
Calculate the concentration of the acid in mol L^{1}.
 What have you been asked to do?
Calculate the concentration of the hydrochloric acid
[HCl_{(aq)}] = ? mol L^{1}
 What information (data) have you been given?
Extract the data from the question:
pH = 3.6
 What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding the concentration of hydrogen ions in solution:
[H^{+}] = 10^{pH}
Substitute in the pH value and solve:
[H^{+}] = 10^{3.6} = 2.5 × 10^{4} mol L^{1}
Write the balanced chemical equation for the dissociation of the acid:
HCl → H^{+}_{(aq)} + Cl^{}_{(aq)}
Find the mole ratio (stoichiometric ratio) :
H^{+} : HCl
1 : 1
 Determine the concentration of the acid using the mole ratio (stoichiometric ratio):
1 mole per litre H^{+} is produced by 1 mole per litre HCl
So, 2.5 × 10^{4} mol L^{1} H^{+} is produced by 2.5 × 10^{4} mol L^{1} HCl
Concentration of the acid is 2.5 × 10^{4} mol L^{1}
 Is your answer plausible?
Use your calculated value for [HCl_{(aq)}] to find pH and compare it to that given in the question:
HCl_{(aq)} → H^{+}_{(aq)} + Cl^{}_{(aq)}
[HCl_{(aq)}] = [H^{+}] = 2.5 × 10^{4} mol L^{1}
pH = log_{10}[H^{+}] = log_{10}[2.5 × 10^{4}] = 3.6
Since this value is the same as that given in the question we are confident our answer is correct.
 State your solution to the problem:
[HCl_{(aq)}] = 2.5 × 10^{4} mol L^{1}
3. 0.25 L of an aqueous solution of hydrochloric acid has a pH of 3.5
Calculate the moles of hydrogen ions present in the solution.
 What have you been asked to do?
Calculate the moles of hydrogen ions
n(H^{+}) = ? mol
 What information (data) have you been given?
Extract the data from the question:
pH = 3.5
volume = V = 0.25 L
 What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding the concentration of hydrogen ions in solution:
[H^{+}] = 10^{pH}
Substitute in the pH value and solve:
[H^{+}] = 10^{3.5} = 3.2 × 10^{4} mol L^{1}
Write the equation (formula) for finding moles given concentration and volume
moles = concentration (mol L^{1}) × volume (L)
concentration of hydrogen ions = [H^{+}] = 3.2 × 10^{4} mol L^{1}
volume of solution = 0.25 L
 Substitute the values into the equation and solve:
moles = concentration (mol L^{1}) × volume (L)
moles(H^{+}) = 3.2 × 10^{4} mol L^{1} × 0.25
= 8.0 × 10^{5} mol
 Is your answer plausible?
Use your calculated value for moles of H^{+} to find pH and compare it to that given in the question:
[H^{+}] = n(H^{+}) ÷ V(solution) = 8.0 × 10^{5} mol ÷ 0.25 L = 3.2 × 10^{4} mol L^{1}
pH = log_{10}[H^{+}] = log_{10}[3.2 × 10^{4}] = 3.5
Since this value is the same as that given in the question we are confident our answer is correct.
 State your solution to the problem:
n(H^{+}) = 8.0 × 10^{5} mol
1. A hydrogen ion is a hydrogen atom that has lost an electron.
Given that the most common naturally occurring isotope of hydrogen contains just 1 proton and NO neutrons in its nucleus, and 1 "orbiting" electron, then, when this isotope loses an electron it is just a proton!
Because naturally occuring hydrogen is also made up of very tiny amount of other isotopes of hydrogen, the term hydron is preferred by IUPAC to represent H^{+}, but most Chemists will still use the term proton when referring to H^{+}.
Most hydrogen ions, H^{+}, are just a naked proton!
A naked proton is very reactive, so, in practice an H^{+} ion "jumps" onto a water molecule to form the hydronium (or oxonium) ion, H_{3}O^{+}.
For this reason H_{3}O^{+} is also known as a hydrated hydrogen ion or hydrated proton.
When Chemists refer to hydrogen ions, hydrons, or H^{+} in aqueous solutions, they really mean H_{3}O^{+}.
Should you write H^{+} or H_{3}O^{+} when talking about aqueous solutions?
Generally speaking, it doesn′t matter, but it would be better to refer to H^{+}_{(aq)}, rather than H^{+}, so that there is no confusion about the nature of the proton.
We use H^{+} and H^{+}_{(aq)} here because it highlights the fact that pH relates to H^{+} concentration.
If you decide to use H_{3}O^{+} instead of H^{+}_{(aq)}, then the equation for finding H_{3}O^{+} concentration becomes:
[H_{3}O^{+}] = 10^{pH}