# 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, HNO**_{3}, HClO_{3}, HClO_{4}and Η_{2}SO_{4}(1^{st}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 C

_{1},C

_{2}, C

_{3},.. 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

STEP #4: From (1)+(9): pH = -log [H

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 V

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

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

V_{FINAL} =
V_{HBr} + V_{HCl} = 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

From (7)+(8) and since HCl and ΗBr are strong acids and
dissociate completely in water:^{-2}mol/ℓ HCl (8)[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 cm*

^{3}of 0.03 M HCl and 100 cm^{3}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 V

_{total}:
V

_{total}= V_{HCl}+ V_{NaOH}=100 cm^{3}+ 100 cm^{3}= 200 cm^{3}(2)The moles of "pure" HCl in the final solution can be calculated as follows:

Volume V =1000 cm

^{3}of HCl solution contains 0.03 moles "pure" HCl
Volume V = 100cm

a

The concentration of [HCl] in the final solution is found to be:^{3}of HCl solution contains a_{1}= ? moles "pure" HCla

_{1}= 3 * 10^{-3}mol HCl
Volume Vtotal= 200 cm

^{3}of the final solution contains 0.03 moles "pure" HCl
Volume Vtotal= 1000 cm

^{3}of the final solution contains a_{2}= ? moles"pure" HCl
a

Similarily, it can be shown that the concentration of NaOH in the final solution
is equal to:_{2}= [HCl] = 0,015 mol (3)
[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

Therefore: [H^{+}and 0.01 MCl^{-}.^{+}] = 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.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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