Weak Acids and Bases – Calculate the pH of a weak acid (a general equation)

The weak acid (base) general equation

Most acids known are weak while there are no more than twenty strong acids. The reactions of weak acids and bases with water do not go to completion. So to calculate the pH of their solutions (pH of a weak acid or pH of a weak base ) the acid (or base) dissociation constants are used, k_a and k_b respectively, and the laws of chemical equilibrium.

Let us now find a general equation for the pH of a weak acid HA (or for [H⁺]). This is the mathematical approach.

Mathematical Approach:

There are two equilibria present: i) the dissociation of water and ii) the dissociation of the weak acid k_a 

[H⁺] [OH^-] = k_w        (1)

[H⁺] [A^-] = k_a [HA]        (2)

If the concentration of the acid in solution is C (M, moles/l), a mass balance on the anion A gives:

C = [HA] + [A^-]            (3)

and a charge balance gives:

[H⁺] = [OH^-] + [A^-]             (4)

Let us assume that k_a, k_w  and C are known. Then there are four equations and four unknowns (shown in red in equations (1) to (4)).

A general expression for [H⁺] (or for the pH of a weak acid) in this case would be an expression in terms of the known k_a, k_w  and C. Therefore, the unknowns have to be eliminated starting from [HA] or [OH^-] that are contained in the least number of equations.

Let us eliminate first [OH^-]. Solving equation (1) for [OH^-] and substituting it in equation (4):

[H⁺] [OH^-] = k_w   and  [OH^-] = k_w / [H⁺]   (1’)

By substituting (1’) to (4) eliminates [OH^-]:

[H⁺] = [OH^-] + [A^-] = k_w / [H⁺] + [A^-]           (5)

Next let us eliminate [HA] by solving equation (3) for [HA] and substituting in equation (2):

C = [HA] + [A^-]   and [HA] = C - [A^-]     (3’)

Substitute (3’) in equation (2) and eliminate [HA]:

[H⁺] [A^-] = k_a [HA]  = k_a (C - [A^-])     (6)

Now let us eliminate [A^-] by solving equation (5) for [A^-] and substitute in equation (6):

[H⁺] = k_w / [H⁺] + [A^-]  and  [A^-] = [H⁺] - k_w / [H⁺]     (7)

Substituting (7) in equation (6):

[H⁺] [A^-] = k_a [HA]  = k_a (C - [A^-])  and

[H⁺] ([H⁺] - k_w / [H⁺])  = k_a (C -[H⁺] - k_w / [H⁺]) )      (8)

Rearranging (8) we get a cubic equation in [H⁺]:

[H⁺]³ + k_a[H⁺]² – (k_w + k_a C) [H⁺] - k_w k_a = 0       (9)

Using the known values of C, k_a, k_w  equation (9) can be solved for [H⁺] and the pH can be calculated. The solution of the cubic equation is straightforward but tedious.

In our days, solution of equations like (9) has become easier by using on line cubic calculators.

Example I.1

Calculate the pH of a 0.20 M solution of HCN (k_a = 4.9*10^-10)

Data	[HCN]= C = 0.20 M ka = 4.9*10-10
Unknown	pH = ?

The k_a indicates that HCN is a weak acid. It dissociates partially in water according to:

By following the above mathematical treatment (exact solution) equation (9) becomes:

[H⁺]³ + (4.9*10^-10)*[H⁺]² – (10^-14+ 4.9*10^-10 *2 * 10^-1)*[H⁺] – (10^-14*4.9*10^-10) =0     

Solutions to equations like the above cubic equation are given by on-line cubic calculators (Fig. I.1):

Fig. I.1: Three solutions to the above cubic equation are given by the Wolfram/Alpha calculator. The positive solution is the one with any physical meaning. Therefore [H⁺] = 9.9 * 10^-6 M.

Therefore [H⁺] = 9.9 * 10^-6 M and the pH of the solution can be calculated:

pH = -log[H⁺] = -log (9.9 * 10^-6) = 5.0

Approximations on the Equation giving [H⁺] for weak acids

 

The error of equilibrium data are usually ± 5% so it is usually possible to make a number of approximations on equation (9) greatly simplifying the solution of problems like the one given in Example I.1:

Case 1

If the acid is not extremely weak or the solution very dilute, the concentration of [OH^-] can be neglected in comparison to that of [H⁺] since [OH^-] <<< [H⁺]. Therefore:

[H⁺] ([H⁺] - k_w / [H⁺])  = k_a (C -[H⁺] - k_w / [H⁺]) )      (8)

[H⁺] ([H⁺] - k_w / [H⁺]) = k_a (C -[H⁺] - k_w / [H⁺]) ) ⇒ [H⁺] ([H⁺] - [OH^-]) = k_a (C -[H⁺] - [OH^-] ) ⇒ [H⁺] [H⁺] = k_a (C -[H⁺])  ⇒

k_a = [H⁺]² / C -[H⁺]      (8a)

Equation (8a) is a quadratic equation:

[H⁺]² + k_a [H⁺] - k_a C = 0       (8a)

Case 2

If the acid is fairly concentrated (more than 10^-3 M) a simplification can be achieved by making the assumption that [H⁺] <<< C since the acid dissociates slightly and very little [H⁺] is produced. Therefore equation (8a) reduces to:

k_a = [H⁺]² / C    ⇒   [H⁺] ≈ (k_a C)^1/2       (8b)

As a general rule:

Several approximations can be made on the mass and charge balances by neglecting a concentration when it is added to another.

The approximate system of equations is solved

The results are checked for consistency with the approximations

In the final form, all approximations should introduce an error of no more than 5

If all the above show that the approximations are not appropriate the equation must be solved in its exact form.

Calculating k_a from pH or pH from k_a

In order to calculate either the k_a for a weak acid (k_a value) when the pH of its solution is known or the pH of a weak acid when the k_a value is known the following procedure is followed (Fig. I.2):

Fig. I.2: Left: Flowchart showing how to calculate the pH of a weak acid solution when k_a and initial concentration C of the acid is known (1-4).  

Right: Flowchart showing how to calculate the k_a  of a weak acid solution when pH and initial concentration C of the acid is known (A to E)

Let’s try to solve now the problem in Example I.1 using the above step-by-step approach – let’s call it chemical approach - and compare with the mathematical approach previously used. The drawback of using the mathematical approach is that formulas have to be memorized that may lead to significant errors since they only give correct results when certain assumptions are fulfilled.

Chemical Approach:

Example I.1 (continued…)

Calculate the pH of a 0.20 M solution of HCN (k_a = 4.9*10^-10)

Given	[HCN]= C = 0.20 M ka = 4.9*10-10
Asked for	pH = ?

Step 1: Write the ionization equilibrium reaction

Step 2: Write the equilibrium constant expression and the value for the equilibrium constant.

k_a = [H⁺] [CN^-] / [HCN] = 4.9*10^-10

Step 3: Assume that x moles/l of HCN dissociate.  Then the concentrations of the species involved in the equilibrium are:

Headspace Analysis - Gas Chromatography - Analysis of Ethanol in Blood (Forensic Chemistry)

Headspace techniques are employed in conjunction with gas chromatography (GC) analysis for certain type of samples. The GC can handle any gaseous sample and any liquid sample that can be vaporized completely and instantaneously before it goes to a proper column for separation. Unfortunately there are many liquid samples (biological, environmental) that cannot be directly injected into the GC. For such samples headspace analysis can be used.

The principle underlying GC headspace analysis is that in a sealed vial at constant temperature equilibrium is established between volatile components of a liquid or solid sample in the vial and the gas phase above it – the “headspace” (Fig. I1). After allowing due time for equilibration (normally 15 min.) a portion of the headspace – ambient volume above a sample matrix where the volatile compounds exist in gaseous form at predictable levels - may be withdrawn via a rubber septum using a gas-tight syringe and injected into the GC column.

Fig. I1: GC headspace vial

Headspace analysis is useful for situations where:

• The analyte of interest is volatile at temperatures below 290 ^∘C

• The sample matrix is a solid, liquid, paste that is not easy to inject into a GC inlet

• Sample preparation to allow easy liquid injection is difficult 

The following video demonstrates the headspace sampler of a GC-FID system:

  Advantages of Headspace Analysis

Headspace analysis provides several advantages over normal injections:

• Simpler sample preparation

• Directly analyze a wide range of sample matrices such as liquids, solids and pastes

• Columns last longer, with less maintenance. The headspace volume above the sample matrix is more clean than the matrix. By injecting fewer contaminants the analytical column lasts longer.

• High precision

• Solvent peak is smaller or nonexistent compared to traditional liquid injection GC techniques.

An internal standard may be added prior to the heating process, and quantitative analyses may be performed after constructing a calibration graph.

This technique is widely used in the analysis of ethanol in blood (blood-alcohol in driving under influence (DUI) and driving whil and other volatile substances in biological samples and in the pharmaceutical industry for measuring solvent residues in tablets, amongst other applications. A gas chromatogram of an in-house reference material containing ethanol, methanol and n-propanol obtained by using headspace analysis is shown in Fig. I2.

Fig. I2: A gas chromatogram of an in-house reference material containing ethanol, methanol and n-propanol obtained by using headspace analysis is shown above. The first peak is acetaldehyde (retention time: 0.886 min), the second peak is ethanol (retention time: 1.407 min) and the last peak is isopropanol the internal standard (retention time: 3.070 min)

Gas Laws – Ideal Gas Law

Gas Laws – Ideal Gases

 

The Main Gas Laws

The three main gas laws are stated below:

Boyle’s law: The volume of a fixed amount of gas maintained at a constant temperature is inversely proportional to the gas pressure:

P ∝ 1/V  or  P = k/V  or P*V = k   (moles n and temperature T constant)  (1)

Equation (1) shows that the product of the pressure and volume of a fixed amount of gas at a constant temperature T is a constant k.

Charles’ law: The volume of a fixed amount of gas at constant pressure is directly proportional to the temperature T (Kelvin)

V ∝ T  or  V = c * T  (where c, pressure P and moles n constant)     (2)

Avogadro’s law: Equal volumes of different gases compared at the same temperature and pressure contain equal numbers of molecules.

V ∝ n  (P and T constant)        (3)

By combining (1), (2) and (3) above into one proportionality:

V ∝ n*T/P     (4)

Proportionality (4) can be replaced by an equality if a proportionality constant R would be included:

V = R*n*T/P   or    P*V = n*R*T     (5)

This proportionality constant is known as the gas constant R.

Any gas that obeys (1), (2) and (3) will also obey equation (5) which is called the ideal gas equation (Fig. I1) . All gases that obey this equation are called ideal gases.

Under suitable conditions some real gases do approach the behavior of ideal gases and make equation (5) very useful.

Fig. I1: Interrelationship of the gas laws. Any gas that obeys (1), (2) and (3) will also obey equation (5) which is called the ideal gas equation

The ideal gas equation can be used to establish molecular weights of gases. For this purpose it is helpful to alter the equation slightly by substituting where n (moles of gas) with its equivalent m/MW (where m is the mass of gas and MW its molecular weight) to get the following equation:

 P*V = (m/MW)*R*T  (5a)

A solved example regarding the determination of the molecular weight of an ideal gas is presented in the following video:

 

 

 

Other Gas Laws

Some other gas laws of note are Raoult’s law, the law of Gaseous Diffusion, Graham’s law and Gay Lussac’s law.

Raoult’s law states: i) the partial pressure of a solute is proportional to the mole fraction of the solute in the solution and ii) the vapor pressure of a solution is directly proportional to the mole fraction of solvent present.

P_soln = x_solvent * Po_solvent        (6)

Where P_soln the observed vapor pressure of the solution

Po_solvent the vapor pressure of the pure solvent

xsolvent is the mole fraction of the solvent in the solution     

From equation (6) can be derived that for a solution that contains half solute molecules and half solvent molecules – xsolvent is 0.5 – the vapor pressure of the solution would be half of the vapor pressure of the solvent.

The effect of the solute on the vapor pressure of a solution gives us a convenient way to “count” molecules and thus provides a means for experimentally determining molar masses. Suppose a certain mass of a compound is dissolved in a solvent, and the vapor pressure of the resulting solution is measured. Using Raoult’s law, we can determine the number of moles of solute present. Since the mass of this number of moles is known, we can calculate the molar mass.

 

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Relevant Posts

Dalton's law - Law of Partial pressures 

Properties of Solutions - Henry's law 

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References

David W. Oxtoby, H.P. Gillis, Alan Campion, “Principles of Modern Chemistry”, Sixth Edition, Thomson Brooks/Cole, 2008

StevenS. Zumdahl, “Chemical Principles”  6thEdition, Houghton Mifflin Company, 2009

Ralph H. Petrucci, “General Chemistry”, 3rd Edition, Macmillan Publishing Co., 1982

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Key Terms

gas ideal gas law, P.V = nRT, the gas laws, Boyle's gas law,Raoult's law, Graham's law, ideal gas law constant, gas law practice problems, Avogadro's, Graham's, Gay Lussac's, gas constant R, chemistry net, ideal gas law equation

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