We’ll show you how to calculate the pH of a solution formed by mixing a strong acid with a strong base. Strong Acid– Strong Base Mixtures

We’ll show you how to calculate the pH of a solution formed by mixing a strong acid with a strong base.

Strong Acid– Strong Base

Mixtures

We’ll consider adding equal amounts of a strong acid and a strong base to the beaker on the bottom.

Strong Acid

Strong

Base

Strong acids contribute hydronium ions to aqueous solutions.

H3O+

Strong Acid

Strong

Base

Acids can more simply be viewed as donating protons or H+ ions. HERE, we’ll consider acids as a source of H+ ions.

H+

Strong Acid

Strong

Base

And bases are a source of hydroxide or OH minus ions.

H+

Strong Acid

Strong

Base

OH–

When a strong acid is added to a container (click), it brings H+ ions with it.

Strong Acid

Strong

Base

OH–H+

And when a strong base is added (click), it brings hydroxide or OH minus ions with it. Of course when the base is added it immediately mixes with the acid.

Strong Acid

Strong

Base

H+

OH–

However, here we’ll pretend it doesn’t mix yet, so we can get a more detailed idea of the processes.

OH–

Strong Acid

Strong

Base

H+

As the solutions mix, H+ and OH minus ions move toward each other.

OH–

Strong Acid

Strong

Base

H+

And in the process of neutralization, react with each other to form water.

OH–

Strong Acid

Strong

Base

H+

H2O

2H OOHH

And the spectator ions in the acid and base will be present as a neutral salt.

Strong Acid

Strong

Base

H2O and a Salt

Now we’ll use a slightly different model to visualize adding a strong acid to a strong base. In this case we’ll start with equal moles (click) of H+ and OH minus, and move these together.

Strong Acid

H+

H+

H+

H+

H+

H+

Strong

Base

OH–

OH–

OH–

OH–

OH–

OH–

The H+ and OH- will neutralize each other to form water.

Strong Acid

H+

H+

H+

H+

H+

H+

Strong

Base

OH–

OH–

OH–

OH–

OH–

OH–

H2OH2OH2OH2OH2OH2O

and there is no strong acid or base left over,

H2OH2OH2OH2OH2OH2O

so the final mixture is neutral

H2OH2OH2OH2OH2OH2O

The Final Mixture

is Neutral

This time the acid we’re starting with has more H+ than the base has OH minus. The H+ is in excess.(click)

Strong Acid

H+

H+

H+

H+

H+

H+

H+

Strong

Base

OH–

OH–

OH–

OH–

OH–

OH–

Each OH- ion will neutralize an H+ ion to form water

Strong Acid

H+

H+

H+

H+

H+

H+

H+

Strong

Base

OH–

OH–

OH–

OH–

OH–

OH–

H2OH2OH2OH2OH2OH2O

But since we had started with an excess of H+, we have some H+ left over. There weren’t enough OH minus ions available to neutralize all the H+ ions we added.

H+

H2OH2OH2OH2OH2OH2O

ExcessH+ Left Over

Because there is H+ left over, the final mixture is Acidic

H+

H2OH2OH2OH2OH2OH2O

The Final Mixture is Acidic

ExcessH+ Left Over

Now we’ll look at another combination.This time the base we’re starting with has More OH minus than the acid has H+. The OH minus is in excess (click). So we combine them.

Strong

Base

OH–

OH–

OH–

OH–

OH–

OH–

Strong Acid

H+

H+

H+

H+

H+

Each H+ ion present will neutralize one OH- ion to form water.

Strong

Base

OH–

OH–

OH–

OH–

OH–

OH–

Strong Acid

H+

H+

H+

H+

H+

H2OH2OH2OH2OH2O

This time, we have excess OH minus left over.

OH–

H2OH2OH2OH2OH2O

Excess OH– Left

Over

so the final mixture is basic.

OH–

H2OH2OH2OH2OH2O

Excess OH– Left

Over

The Final Mixture is Basic

Now we’ll do an example question. We’re told that 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH.

18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH.

And we’re asked to find the pH of the final mixture.

18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

We’ll begin by finding the initial moles of H+ added.


0.200 molmol HCl 0.0180 L 0.00360 mol H

1m

Lol Hin ini itial ltia

0.160 mol mol OH mol NaOH 0.0250 L 0.00400 mol OH

1Linitial initial

excess mol OH 0.00400 mol OH 0.00360 mol H 0.00040 mol OH

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L

pOH log OH log 0.0093 2.03 pH 14.00 pOH 14.00 2.03 11.97

Because the source of H+, HCl is a strong monoprotic acid, the moles of H+ is equal to the moles of HCl.


0.200 molmol H 0.0180 L 0.mol 00360 mol H

1L HClin ini itial tial


1Linitial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


The moles of HCl is equal to 0.200 moles per L


0.200 mmol H mol HCl 0.00360 mol

ol0.0180 L

1LHinitial initial


1Linitial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


times 0.0180 L


0.200 molmol H mol HCl 0.00360 mol H

10.0180

LLinitial initial


1Linitial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


Which is 0.00360 moles. So 0.00360 moles of H+ was added.


0.200 m0.00360 mol H

olmol H mol HCl 0.0180 L

1Linitial initial


1Linitial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


Now we’ll calculate the initial moles of OH minus added.


0.200 molmol H mol HCl 0.0180 L 0.00360 mol H

1Linitial initial

0.160 mol mol NaOH 0.0250 L 0.00400 mol OH

1H

Lmol O i inn iitial tial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


Because NaOH is a strong base with one OH in its formula, the moles of OH minus is equal to the moles of NaOH added.



1Linitial initial

0.160 mol mol OH 0.0250 L 0.0mo 0400 mol OH

1N

Ll aOHi inn iiti ta iall


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


the moles of NaOH is 0.160 moles per L



1Linitial initial

0.160 mmol OH mol NaOH 0.00400 mol OH

ol 0.0250 L

1Linitial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


times 0.0250 L



1Linitial initial

0.160 mol mol OH mol NaOH 0.00.0250 0400 mol OH

1L

Linitial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


which is 0.00400 moles. So the initial moles of OH minus added is 0.00400 mol.



1Linitial initial

0.160 m0.00400 mol OH

ol mol OH mol NaOH 0.0250 L

1Linitial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


Notice that both the moles of H+ and OH minus are expressed to 3 significant figures, which is consistent with the given data



1L360initial initial


1L400 initial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


Also notice that both of these values have 5 decimal places. Next we check whether moles of H+ or moles of OH minus is in excess


0.200 molmol H mol HCl 0.0180 L 0. mol H

10

L0360initial initial

0.160 m0

ol mol OH mol NaOH 0.0250 L 0. mol OH

1L0400 initial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


0.00400 moles is greater than 0.00360 moles, so OH minus is in excess.



1Linitial initial

0.160 m0.00400 mol OH


1Linitial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


OH– is in Excess

Now, 0.00360 moles of H+ will neutralize 0.0036 moles of OH minus.


0.200 m0.00360 mol H


1Linitial initial


1Linitial initial


0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


Will neutralize 0.00360 mol OH–

So the excess moles of OH minus left over will be



1Linitial initial


1Linitial initial

0.00400 mol OH 0.00360excess m mol H 0.00040 mol Ol HOH o

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


Will neutralize 0.00360 mol OH–

The initial 0.00400 moles



1Linitial initial

0.160 m0.00400 mol OH


1Linitial initial

0.00400 mol Oexcess mol OH 0.00360 mol H 0.00040 mol OHH

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L


minus the 0.00360 moles of H+ that was able to neutralize 0.00360 moles of OH minus.


0.200 m0.00360 mol H


1Linitial initial

excess mol OH 0.00400 mol O 0.00360 moH 0.00040 mol Hl H O

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L



1Linitial initial

Which is equal to 0.00040 moles. And because OH minus is the ion in excess, this is 0.00040 moles of OH minus.



1Linitial initial

excess mol OH 0.00400 mol OH 0.00360 mo 0.00040 mol OHl H

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L



1Linitial initial

Because the numbers that were subtracted to get this both have 5 decimal places, the value 0.00040 must also have 5 decimal places.



1Linitial initial

excess mol OH mol O0.00400 0.00360H mol H mol 0.00 OH040

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L



1Linitial initial

5 decimal places

5 decimal places

5 decimal places

But notice this value, expressed to 5 decimal places, has only 2 significant figures. Zero’s to the left of the first non-zero digit are not significant. So our final answer cannot have more than 2 significant figures.



1Linitial initial

excess mol OH 0.00400 mol OH 0.00360 mol H 0.000 mol OH40

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L



1Linitial initial

2 significant figures

Because OH minus is the ion in excess, we now calculate the concentration of OH minus in the final mixture.



1Linitial initial


0.00040 mol OH 0.00040 mol OH

0.0093 M0.0430 L0.0180 L 0.

Final O0250 L

H



1Linitial initial

Concentration of OH minus is moles of OH minus over the total volume in Litres. We have 0.00040 moles of OH minus



1Linitial initial

excess mol OH 0.00400 mol OH 0.00360 mo 0.00040 mol OHl H

0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.

0.00040 m

0250

ol OH

L



1Linitial initial

And the total volume of the mixture is 0.0180 L of the HCl solution


0.200 molmol H mol HCl 0.00360 mol H

10.0180

LLinitial initial

0.0180 L 0.0250

0.00040 mol OH 0.00040 mol OHFinal OH 0.0093 M

0.0430 LL



1Linitial initial


plus 0.0250 L of NaOH solution.



1Linitial initial

0.0180 L 0.0250

0.00040 mol OH 0.00040 mol OHFinal OH 0.0093 M

0.0430 LL


0.160 mol mol OH mol NaOH 0.00.0250 0400 mol OH

1L

Linitial initial


Or 0.0430 L. So the final concentration of OH minus is 0.00040 moles divided by 0.0430 L,



1Linitial initial

0.00040 mol OH

Final OH 0.0093 M0.0180 L 0.

0.00040 mol OH

0.0430 L0250 L



1Linitial initial


Which is 0.0093 M.



1Linitial initial

0.00040 mol OH 0.00040 mol OH

Final OH0.0430 L0.0180 L 0.

0.0093 M0250 L



1Linitial initial


We want to find the pH of the mixture, but since we have the OH minus concentration, we start by calculating the pOH, which is the negative log of the hydroxide ion concentration



1Linitial initial

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L

log 0.0093 2.pOH log H 03O pH 14.00 pOH 14.00 2.03 11.97


1Linitial initial


Or the negative log of 0.0093



1Linitial initial

0.00040 mol OH 0.00040 mol OH

Final OH M0.0430 L0.0180 L 0.0

0.0025 L

90

3

log 0.00pOH log OH 2.093 3 pH 14.00 pOH 14.00 2.03 11.97


1Linitial initial


Which is 2.03. Take a moment to check this on your calculator. Remember, 2 decimal places in a pOH value is 2 significant figures.



1Linitial initial

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L



1Linitial initial


In the last step, we’ll calculate the pH



1Linitial initial

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L

pOH log OH log 0.0093 2.03 14.0pH 14 0 2.03 11.0 70 p 9OH .


1Linitial initial


Which is 14 minus the pOH



1Linitial initial

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L

pOH log OH log 0.0093 2.03 pH 14.0014.00 2.03 11.97pOH


1Linitial initial


Or 14 minus the 2.03



1Linitial initial

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L

pOH log OH log 0.00 2.0393 14.0pH 14.00 pOH 11.0 92.03 7


1Linitial initial


Which is 11.97.


1Linitial initial

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L

pOH log OH log 0.0093 2.03 14.00 pOH 14.00 2.03pH 11.97


1Linitial initial


pH of the final mixture


So we’ve now answered what we set out to answer. The pH of the final mixture is 11.97. This is expressed to 2 decimal places, which in a pH value, is 2 significant figures.


1Linitial initial

0.00040 mol OH 0.00040 mol OH

Final OH 0.0093 M0.0430 L0.0180 L 0.0250 L

pOH log OH log 0.0093 2.03 14.00 pOH 14.00 2.03pH 11.97


1Linitial initial


pH of the final mixture


Documents

We’ll show you how to calculate the pH of a solution formed by mixing a strong acid with a strong base. Strong Acid– Strong Base Mixtures