Here, we’ll show you how to calculate the initial concentration of a weak acid, given the pH and...

Here, we’ll show you how to calculate the initial concentration of a weak acid, given the pH and the Ka of the acid. In this particular example, we’ll use the initial concentration to find initial moles of the acid in the sample

pH and Ka to Initial

Concentration

We’re told that we have a 500.0 mL sample of oxalic acid (H2C2O4) with a pH of 1.095.

A 500.0 mL sample of oxalic acid (H2C2O4) has a pH of 1.095.

And we’re asked how many moles of H2C2O4 were initially added to this sample?

A 500.0 mL sample of oxalic acid (H2C2O4) has a pH of 1.095. How many moles of H2C2O4 were initially added to the sample?

We’ll divide the solution to this problem into 6 main steps.


Six Steps:

We’re given the pH, and since oxalic acid is a weak acid, we will need an ICE table. We can use hydronium concentration in an ICE table, but not pH. So step 1 is to convert the pH to hydronium ion concentration.

Six Steps:1. Convert pH to [H3O+]


In step 2, we write the balanced ionization equation for oxalic acid.

Six Steps:1. Convert pH to [H3O+]2. Write the ionization equation for oxalic acid


Next, we draw an ICE table below the equation

Six Steps:1. Convert pH to [H3O+]2. Write the ionization equation for oxalic acid3. Draw an ICE table below the equation


We let x equal the initial concentration of the acid, insert known quantities into the ICE table, and determine all the other quantities.

Six Steps:1. Convert pH to [H3O+]2. Write the ionization equation for oxalic acid3. Draw an ICE table below the equation4. Let x equal the initial concentration of oxalic acid,

insert known quantities into the ICE table, and determine the other quantities


We write the Ka expression for oxalic acid, look up the Ka value, insert equilibrium concentrations into the expression, and solve for x, the initial concentration



5. Write the Ka expression for oxalic acid, look up the Ka value, insert equilibrium concentrations into the Ka expression, and solve for x, the initial concentration


In the last step, we’ll use the concentration and volume of the sample to calculate the number of moles of oxalic acid initially added to the sample.




6. Using the concentration and volume, calculate the number of moles of oxalic acid initially added


We’ll start the process by converting the given pH to hydronium ion concentration.

1. Convert pH to [H3O+]


We use the equation hydronium concentration equals 10 to the negative pH


1.095

p

3

H3

3 10

H O 0.080

H O 1

O

4 M

0

H


Which is 10 to the negative 1.095


1.0

H

95

p3

3

3

H O 10

H O 10

H O 0.0804 M


This works out to 0.0804 molar


1.0

H

95

p3

3

3

H O 10

H O 10

H O 0.0804 M


So we can state that the hydronium concentration is 0.0804 molar


pH3

1.0953

3

H O 10

H O 10

H O 0.0804 M


And we’ll make a note of it up here


pH3

1.0953

3

H O 10

H O 10

H O 0.0804 M

[H3O+] = 0.0804 M


In step 2, we write out the ionization equation for oxalic acid.

2. Write the ionization equation for oxalic acid, H2C2O4.

[H3O+] = 0.0804 M


We add H2C2O4 to water


[H3O+] = 0.0804 M

2 2 4(aq) 2 ( ) 3 (aq) 2 4(aq)H C O H O H O HC Ol


And it forms H3O plus,


[H3O+] = 0.0804 M



And the conjugate base of oxalic acid, the hydrogen oxalate ion, HC2O4 minus


[H3O+] = 0.0804 M



The hydrogen oxalate ion

Now, we draw an ICE table below the equation and add a grid so that the columns line up with the substances in the equation.

3. Draw an ICE table [H3O+] = 0.0804 M

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804



Water is a liquid, so we ignore its values. We’ll colour its column blue here.


[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804



We’ll start by adding what we can to the initial concentration row.


[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804



We don’t know what the initial concentration of oxalic acid is, so we let it equal


[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804


?


x

4. Let x equal the initial concentration of oxalic acid, insert known quantities into the ICE table, and determine the other quantities

[H3O+] = 0.0804 M

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804



And before ionization, the concentrations of hydronium and hydrogen oxalate ions are zero.

[H3O+] = 0.0804 M

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




Remember, using the pH, we had determined that the hydronium concentration at equilibrium is 0.0804 molar

[H3O+] = 0.0804 M


So we’ll write 0.0804 in the cell for the equilibrium concentration of hydronium, down here.

[H3O+] = 0.0804 M

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




Now we’ll focus on the column for hydronium.

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




The hydronium ion concentration started out as 0 molar and went up to 0.0804 molar at equilibrium, so its change in concentration is an increase of 0.0804 molar.

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




Hydronium and hydrogen oxalate both have a coefficient of 1 in the ionization equation.

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




1 1

So we can say that the concentration of hydrogen oxalate also increased by 0.0804 molar

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




1 1

Because concentrations of the products increased, we can see that the reaction (click) has moved to the right in order to establish equilibrium

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804



4. Let x equal the initial concentration of oxalic acid, insert known quantities into the ICE table, and determine the other quantities Moved to the Right

Therefore, we can say that the concentration of oxalic acid on the left must have decreased.

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804



4. Let x equal the initial concentration of oxalic acid, insert known quantities into the ICE table, and determine the other quantities Moved to the Right

And because the coefficients on oxalic acid and hydronium are both 1

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




11

We can say that the concentration of oxalic acid went down by 0.0804 molar, the same as the hydronium ion increased by.

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




11

Now we can fill in the row for equilibrium concentrations.

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




We’ll start with the hydrogen oxalate ion on the right. It’s concentration started out as 0 molar

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




And increased by 0.0804 molar

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




So its equilibrium concentration is 0.0804 molar

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




Now we’ll move over to the oxalic acid on the left. It’s concentration started out as x molar

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




And it went down by 0.0804 molar

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




So its equilibrium concentration is x minus 0.0804 molar

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




So now we have equilibrium concentrations for all three species.

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804




Our task now is to determine the value of x.

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804A 500.0 mL sample of oxalic acid (H2C2O4) has a pH of

1.095. How many moles of H2C2O4 were initially added to the sample?


?

?


We do this by inserting these equilibrium concentrations into the Ka expression for oxalic acid as set out in Step 5 of our plan.

[I] x 0 0[C]

– 0.0804+

0.0804 + 0.0804

[E] x – 0.0804 0.0804 0.0804A 500.0 mL sample of oxalic acid (H2C2O4) has a pH of

1.095. How many moles of H2C2O4 were initially added to the sample?



We start by using the ionization equation to write the Ka expression for oxalic acid. It is products over reactants,




3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

Which is the concentration of hydronium times the concentration of hydrogen oxalate over the concentration of oxalic acid.




3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

Looking up oxalic acid on the acid table, it tells us that it’s Ka value is 5.9 × 10-2




3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

To save space, we’ll write this as a decimal number. 5.9 × 10-2 is the same as 0.059



3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

Ka = 0.059

We can now substitute this value for Ka in our Ka expression.



a

3 2 4

2 2 4

2

2 2 4

2

3 2 4

2 2 4

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.080

H O HC O

H C O

4

K

x

Ka = 0.059

Like this



3 2 4

2 2 4

3 2 4

2

2

2

2 4

2

2 4

a

0.059

0.08040.059

H C O

0.08040.059

0.0804

H O HC O

H C O

H O HC O

H C O

K

x

Ka = 0.059

We substitute 0.0804 in for both hydronium and hydrogen oxalate, and their product in the Ka expression is 0.0804 squared.



3 2 4

3 2 4a

2 2 4

2 2 4

2 2 4

2

2

H O H

H O HC O

H C O

0.059H C O

0.0590.0804

0.0

C

8040.059

0.0

H

8 4

C

0

O

O

K

x

And we’ll substitute x - 0.0804 in for the equilibrium concentration of H2C2O4.



3 2 4a

2 2 4

3 2 4

2

2 4

2

2 4

2

2

0.080

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

0.08040.059

H C

4

O

K

x

So now we have the equation 0.059 equals 0.0804 squared over x minus 0.0804.



3 2 4a

2 2 4

3 2 4

2

2

2

2 4

2

2 4

0.08040.059

0.080

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C

4

O

K

x

We’ll rearrange this equation to solve for x minus 0.0804, and we get x minus 0.0804 equals 0.0804 squared over 0.059



3 2 4a

2 2 4

3 2 4

2

2

2

2 4

2

2 4

0.08040.059

0.080

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C

4

O

K

x

2

2 2 4

0.0804 0.1096

0.1096 0.0804

0.08040.0804

0.0

0.19 M

H C O 0.19

59

Minitial

x

x

x

x

0.0804 squared over 0.059 works out to 0.1096. We’ll leave this as 4 significant figures for now and round our answer to the correct number of significant figures at the end of the calculation.



3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

2 2 4

2

0.0804

0.0804

0.1096 0.08

0.0804

0.00

04

0.1

.10

9 M

59

H C O 0.19 M

96

initial

x

x

x

x

So at this point, we know that x minus 0.0804 is equal to 0.1096 .



3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

2

2 2 4

0.08040.0804

0.059

0.1096 0.0804

0.0804 0.10

0.19 M

H C O 0.19 M

96

initial

x

x

x

x

Adding 0.0804 to both sides of the equation, gives us x is equal to 0.1096 plus 0.0804



3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

2

2 2 4

0.109

0.08040.0804

0.059

0.19 M

H C O 0.19

0.0804 0

6 0.

.1096

M

0804

initial

x

x

x

x

Which is 0.190.



3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

2

2 2 4

0.1096 0

0.

.0

0804

804

0.08040.059

0.0804 0.1096

H C O 0.19

0.190

0 Minitial

x

x

x

x

Remember that x was set as the initial concentration of oxalic acid, H2C2O4.



3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

2 2 4

20.0804

0.08040.059

0.0804 0.1096

0.1

0.

H C O 0.190 M

096 0.0804

190

initial

x

x

x

x

So we can say that the initial concentration of H2C2O4 was 0.190 molar. We added the unit M because this is a molar concentration.



3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

2 2 4

20.0804

0.08040.059

0.0804 0.1096

0.1

0.

H C O 0.190 M

096 0.0804

190

initial

x

x

x

x

We’ll make a note of it up here.



3 2 4a

2 2 4

3 2 4

2 2 4

2

2 2 4

2

H O HC O

H C O

H O HC O0.059

H C O

0.08040.059

H C O

0.08040.059

0.0804

K

x

2 2 4

20.0804

0.08040.059

0.0804 0.1096

0.1096 0.0804

0.190

H C O 0.190 Minitial

x

x

x

x

[H2C2O4]initial

= 0.190 M

We now know the initial concentration of the oxalic acid.


[H2C2O4]initial

= 0.190 M

We know the initial concentration

But the question asks us to find the number of moles of Oxalic acid.


[H2C2O4]initial

= 0.190 M

We know the initial concentration

So in Step 6, we’ll use the concentration and volume to calculate the number of moles of oxalic acid initially added to the sample.



2 2 4

2 2 4

2 2 4 2

4

2

2 2

4

0.190 mol H C O0.5000 L

1Lmol

H C O

mol H C O 0.095 mol

H C

H O

O

C

[H2C2O4]initial

= 0.190 M

The moles of H2C2O4



[H2C2O4]initial

= 0.190 M

2 2 4

2 2 4

2 2 4 2

4

2

2 2

4

0.190 mol H C O0.5000 L

1Lmol

H C O

mol H C O 0.095 mol

H C

H O

O

C

Is equal to 0.190 moles of H2C2O4 per litre



[H2C2O4]initial

= 0.190 M

2 2 4

2 2 42 2 4

2 2 4 2 2 4

mol H C O 0.5000 L

mol H C O 0.095 mol H C O

0.190 mol H C O

1 L H C O

Times the volume of the sample, 0.5000 L



2 2 42 2 4

2 2 4

2 2 4 2 2 4

0.5000 L0.190 mol H C O

mol H C O1 L H C O


Equals 0.095 moles of H2C2O4.



2 2 42 2 4

2 2 4

2 2 4 2 2 4

0.190 mol H C Omol H C O 0.5000 L

1 L

mol H C O 0.095 mol

H C

H O

O

C

So we can say that the number of moles of H2C2O4 initially added to the sample was 0.095 moles.



2 2 42 2 4

2 2 4

2 2 4 2 2 4

0.190 mol H C Omol H C O 0.5000 L

1 L

mol H C O 0.095 mol

H C

H O

O

C

This final answer is rounded to 2 significant figures.



2 2 42 2 4

2 2 4

2 2 4 2 2 4

0.190 mol H C Omol H C O 0.5000 L

1 L H C O

mol H C O 0.0 mol95 H C O

2 significant figures

This is because the Ka value we obtained from the acid table was only 2 significant figures.



2 2 42 2 4

2 2 4

2 2 4 2 2 4

0.190 mol H C Omol H C O 0.5000 L

1 L H C O



2 is the lowest number of significant figures in the data we used, so we round our final answer to 2 significant figures.



2 2 42 2 4

2 2 4

2 2 4 2 2 4

0.190 mol H C Omol H C O 0.5000 L

1 L

mol H C O 0.095 mol

H C

H O

O

C


So to summarize, the final answer is 0.095 moles of oxalic acid, H2C2O4 was initially added to the 500.0 mL sample.



2 2 4 2 2 4mol H C O 0.095 mol H C O

0.095 moles of H2C2O4 was initially added to the

500.0 mL sample.

Here, we’ll show you how to calculate the initial concentration of a weak acid, given the pH and...

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Transcript of Here, we’ll show you how to calculate the initial concentration of a weak acid, given the pH and...