Strong Acids & Bases: Calculations involving mixtures of strong acids and bases
Solutions containing two or more strong acids or two or more strong bases or a strong acid and a strong base are encountered very frequently in many practical applications of chemistry or in the chemical laboratory. In most cases like these chemistry intuition is needed to calculate the pH or pOH values of the solutions.
In previous posts entitled "Strong acid and bases - Weak acid and bases - Dissociation constants and pK's " and "Ionic Equilibrium - Strong Acids and Bases calculation of the pH of a strong acid. " the definition of strong acids and bases has been given. Strong acids and bases dissociate completely in water - and methods to calculate the pH of a strong acid or the pH of a strong base solution has been presented.
The following are the most commom strong acids and bases:
- Salts and hydroxides of group I and II metals of the periodic table (ΝaOH, KOH, Ca(OH)2)
- HCl, HBr, HI, HNO3, HClO3, HClO4 and Η2SO4 (1st dissociation)
Let us examine two examples where mixture of strong acids or strong acids and strong bases are given and the pH of their solution has to be calculated:
EXAMPLE #1
Solutions that contain two or more strong acids or bases with known initial concentrations C1,C2, C3,.. and the pH or pOH or [H+] or [OH-] are the unknowns and have to be calculated.The following 4-step method can be used:
- Write down the data given and the unknowns
- Write chemical reactions (dissociation reactions of the strong base or strong acid in this case ) and equations that relate data given and unknown (s) so that you have equal number of unknowns and equations
- Determine the total [H+]TOTAL = [H+] ACID 1+ [H+] ACID 2 + ... of the solution in equilibrium
- Calculate from the above steps pH, pOH, [H+] or [OH-]
Calculate the pH of a solution that was formed by mixing 10 ml 0.1 M HBr and 20 ml 0.1 Μ HCl
SOLUTION
STEP #2: Write chemical reactions (dissociation reactions of the strong base or strong acid in this case ) and equations that relate data given and unknown(s) so that you have equal number of unknowns and equations
Equations that relate data given and unknowns are the following:
pH = -log [H+] (1)
In order to calculate the pH value in (1) we have to calculate [H+] in the solution after the mixing of the two solutions. The two strong acids completely dissociate to produce H+.
The volume V of the solution after mixing is found to be:
The moles of «pure» ΗΒr dissolved in 10 ml 0.1M HBr solution is found to be:
The moles of «pure» ΗCl in 20 ml 0.15M ΗCl solution is found tobe:
STEP #3: The [ΗBr]concentration after mixing - in the final solution with volume V= 30 ml - is found to be:
The [ΗCl] concentration in the final solution - after mixing the two acidic solutions - with volume V = 30 ml is found to be:
[H+] = 3.3 * 10-2 mol/L +6.6 * 10-2 mol/L = 9.9 * 10-2 mol/L ≈ 0.1M(9)
STEP #4: From (1)+(9): pH = -log [H+] = -log (0.1) ≈ 1
STEP #1: Write down the data given and the unknowns:
STEP #2: Write chemical reactions(the acid-base neutralization reaction in this case ) and equations that relate data given and unknown(s) so that you have equal number of unknowns and equations:
pH = -log [H+] (1)
After mixing the total volume of the solution is equal to Vtotal:
The moles of "pure" HCl in the final solution can be calculated as follows:
The neutralization reaction that takes place and the equilibrium concentrations of the species involved after reaction is shown below:
STEP #4: From (1) + (6): pH = -log [H+] = -log (0.01) = 2
J.N. Butler “Ionic Equilibrium – Solubility and pH calculations”, Wiley – Interscience, 1998
Clayden, Greeves, Waren and Wothers “Organic Chemistry”,Oxford,
D. Harvey, “Modern Analytical Chemistry”, McGraw-Hill Companies Inc., 2000
VFINAL = VHBr + VHCl = 10 ml + 20 ml = 30 ml (2)
The moles of «pure» ΗΒr dissolved in 10 ml 0.1M HBr solution is found to be:
Volume 1000 ml of ΗΒr solution contains 0.1 mol «pure» HBr
Volume 10 ml of ΗΒr solution contain x =; mol «pure» HBr
x = (0.1 mol HBr) * (10 ml) / 1000 ml = 10-3 mol HBr (5)
The moles of «pure» ΗCl in 20 ml 0.15M ΗCl solution is found tobe:
y = (0.1 mol HCl) * (20 ml) / 1000 ml = 2*10-3 mol HCl (6)
STEP #3: The [ΗBr]concentration after mixing - in the final solution with volume V= 30 ml - is found to be:
Volume 30 ml of the final solution contains 10-3 mol HBr
Volume 1000 ml of the final solution contain z =;
z = 3.3 * 10-2 mol/L HBr (7)
The [ΗCl] concentration in the final solution - after mixing the two acidic solutions - with volume V = 30 ml is found to be:
Volume 30 ml of the final solution contain 2*10-3 mol HCl
Volume 1000 ml of the final solution contain n =;
n = 6.6 * 10-2 mol/ℓ HCl (8)
From (7)+(8) and since HCl and ΗBr are strong acids and dissociate completely in water:[H+] = 3.3 * 10-2 mol/L +6.6 * 10-2 mol/L = 9.9 * 10-2 mol/L ≈ 0.1M(9)
STEP #4: From (1)+(9): pH = -log [H+] = -log (0.1) ≈ 1
EXAMPLE #2
What is the pH value of a solution that is prepared by mixing 100 cm3 of 0.03 M HCl and 100 cm3 of 0.01 M NaOH.The total volume remains constant during the process.
SOLUTION
STEP #1: Write down the data given and the unknowns:
STEP #2: Write chemical reactions(the acid-base neutralization reaction in this case ) and equations that relate data given and unknown(s) so that you have equal number of unknowns and equations:
pH = -log [H+] (1)
After mixing the total volume of the solution is equal to Vtotal:
Vtotal = VHCl + VNaOH =100 cm3 + 100 cm3 = 200 cm3 (2)
The moles of "pure" HCl in the final solution can be calculated as follows:
Volume V =1000 cm3 of HCl solution contains 0.03 moles "pure" HCl
Volume V = 100cm3 of HCl solution contains a1 = ? moles "pure" HCl
a1 = 3 * 10-3 mol HCl
The concentration of [HCl] in the final solution is found to be:a1 = 3 * 10-3 mol HCl
Volume Vtotal= 200 cm3 of the final solution contains 0.03 moles "pure" HCl
Volume Vtotal= 1000 cm3 of the final solution contains a2 = ? moles"pure" HCl
a2= [HCl] = 0,015 mol (3)
Similarily, it can be shown that the concentration of NaOH in the final solution is equal to: [ΝaOH] = 0,005 M (4)
The neutralization reaction that takes place and the equilibrium concentrations of the species involved after reaction is shown below:
STEP#3: The concentrations of the species involved at equilibrium are: [HCl]= 0.01 M and [NaCl] = 0.005 M. The pH of the solution is not affected by the ionization of NaCl. The strong acid HCl dissociates completely according to reaction (3) in the previous example to produce 0.01 M H+ and 0.01 MCl-.
Therefore: [H+] = 0.01 M(6)STEP #4: From (1) + (6): pH = -log [H+] = -log (0.01) = 2
Relevant Posts
Strong Acids and Bases – Ionic Equilibrium – A general relation for the pH of a strong acid
Chemical Equilibrium Calculations in Analytical Chemistry
pH of a strong acid – Examples
References
J-L. Burgot “Ionic Equilibria in Analytical Chemistry”, Springer Science & Business Media, 2012J.N. Butler “Ionic Equilibrium – Solubility and pH calculations”, Wiley – Interscience, 1998
Clayden, Greeves, Waren and Wothers “Organic Chemistry”,Oxford,
D. Harvey, “Modern Analytical Chemistry”, McGraw-Hill Companies Inc., 2000
Key Terms
strong acids and strong bases, ph of mixture of strong acid and strong base, calculate pH of a mixture of strong acid and strong base, pH of mixture of strong acids, solutions that contain two or more strong acids or bases, mixture of strong acid and strong base
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