Category: Intermediate Chemistry

Posted on

Start point of titration

The start point of a titration marks the initial pH of the solution before any titrant is added, serving as a crucial reference for analysing the subsequent changes in acidity or alkalinity throughout the process.

To characterised the start point, we begin with the following assumptions:

    • [H+] from water is negligible, i.e. H+ in the flask containing the weak acid is solely due to that of the acid, [H+]a, as the dissociation of water is suppressed at this stage.
    • For a weak acid, [HA]  is approximately equal to the concentration of the undissociated acid, [HA]ud, i.e. the dissociation of the weak acid HA is negligible.

The equation for the dissociation of a weak monoprotic acid is:

HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)

with the equilibrium constant at the start of the titration as:

K_a=\frac{[H^+][A^-]}{[HA]}\approx \frac{[H^+]_a[A^-]}{[HA]_{ud}}=\frac{[H^+]_a\, ^2}{[HA]_{ud}}

Taking the logarithm on both sides of the above equation and rearranging, we have:

pH=\frac{pK_a-log[HA]_{ud}}{2}\; \; \; \; \; \; \; \; (1)

Eq1 is the general formula for determining the pH of a strong base to weak acid titration at the start point.

The second assumption becomes less valid when the weak monoprotic acid or weak monoprotic base has Ka ≥ 10-3 or Kb ≥ 10-3 respectively. Removing the second assumption, the equilibrium constant expression becomes:

K_a\approx \frac{[H^+]_a[A^-]}{[HA]_{ud}-[H^+]_a}=\frac{[H^+]_a\, ^2}{[HA]_{ud}-[H^+]_a}

The corresponding pH equation is obtained by rearranging the above equation into a quadratic equation in terms of  [H+]a, finding the latter’s roots and taking the logarithm of the root:

pH=-log\left ( \frac{\sqrt{K_a\, ^2+4K_a[HA]_{ud}}-K_a}{2} \right )

 

Question

Do both the above equation and eq1 give the same pH value for the titration of 10 cm3 of 0.200 M of CH3COOH (Ka = 1.75 x 10-5) with NaOH?

Answer

Yes, both formulae give pH = 2.73. This validates the applicability of eq1 for acids with Ka in the region of 10-5.

 

If we disregard both assumptions, we will end up deriving the complete pH titration curve for a strong base to weak acid titration. See this article in the advanced section for details.

 

Question

Show that pKa + pKb = 14.

Answer

HA(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+A^-(aq)

A^-(aq)+H_2O(l)\rightleftharpoons HA(aq)+OH^-(aq)

 

K_a=\frac{[H_3O^+][A^-]}{[HA]}\; \; \; \; \; \; \; \; K_b=\frac{[HA][OH^-]}{[A^-]}

-logK_a-logK_b=-log\frac{[H_3O^+][A^-]}{[HA]}-log\frac{[HA][OH^-]}{[A^-]}

pK_a+pK_b=-log[H_3O^+][OH^-]=pK_w=14

 

next article: After start point and before stoichiometric point of titration
Previous article: overview
Content page of strong base to weak acid titration curve
Content page of intermediate chemistry
Main content page
Posted on

Strong base to weak acid titration curve: combining all equations

Equations describing the various stages of a titration provide a mathematical framework to quantify changes in pH of a solution, effectively illustrating the transition from the start point to the endpoint of the titration.

In summary, the pH curve of a strong base to weak acid titration can be described by the following simplified equations in place of a complex complete pH titration equation:

Start point pH=\frac{pK_a-log[HA]_{ud}}{2}\; \; \; \; \; \; \; \; (1)
After start point but before stoichiometric point pH=pK_a+log\frac{[A^-]_s}{[HA]_r}\; \; \; \; \; \; \; \; (2)
Maximum buffer capacity point pH=pK_a\; \; \; \; \; \; \; \; (3)
Stoichiometric point pH=\frac{pK_w+pK_a+log[A^-]}{2}\; \; \; \; \; \; \; \; (4)
Beyond stoichiometric point pH=pK_w+log[OH^-]_{ex}\; \; \; \; \; \; \; \; (5)

The diagram below shows the plot of the above equations on a single graph for the titration of 10 cm3 of 0.200 M of CH3COOH (Ka = 1.75 x 10-5) with 0.100 M of NaOH:

A complementary set of equations can be derived using the same logic mentioned in the earlier articles to describe a strong acid to weak base titration.

next article: Improved Henderson-Hasselbalch equation
Previous article: Beyond stoichiometric point of titration
Content page of strong base to weak acid titration curve
Content page of intermediate chemistry
Main content page
Posted on

pH indicators: overview

A pH indicator is usually a large, weak organic acid that is soluble in water or alcohol.

\begin{matrix} HIn(aq)\\colourA \end{matrix}\rightleftharpoons H^+(aq)+\begin{matrix} In^-(aq)\\colourB \end{matrix}

The acid HIn absorbs a certain range of wavelengths of visible light and reflects the rest (complementary range of wavelengths) into our eyes, which perceive the complementary wavelengths as Colour A. The conjugate base In– absorbs a different range of wavelengths of visible light and reflects a dissimilar complementary range into our eyes, which perceive it as Colour B.

The acid may be a neutral or charged molecule, for example, the acid form of the indicator bromocresol green is monoanionic (yellow), while its conjugate base is dianionic (blue).

 

 

Next article: The pH range of an indicator
Content page of ph indicators
Content page of Intermediate chemistry
Main content page
Posted on

After start point and before stoichiometric point of titration

How can we mathematically describe the titration curve of a weak monoprotic acid versus a strong monoprotic base, after the start point and before the stoichiometric point?

To do so, we will make three assumptions:

    • [H+] from water is negligible, i.e. H+ in the flask containing the weak acid is solely due to that of the acid, [H+]a, as the dissociation of water is again suppressed at this stage.
    • For a weak acid, [HA] is approximately equal to the concentration of the undissociated acid, [HA]ud, i.e. the dissociation of the weak acid HA is negligible and the remaining concentration of HA after the addition of base is [HA]r = [HA]ud – [A–]s.
    • [A–] is approximately equal to the concentration of the salt, [A–]s, in the solution, i.e. A– is completely attributed to the salt formed, with no contribution from the further dissociation of HA.

The equation for the dissociation of a weak monoprotic acid is

HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)

with the equilibrium constant after start point and before stoichiometric point  being:

K_a=\frac{[H^+]_a[A^-]_s}{[HA]_r}

Taking the logarithm on both sides and apply the definitions of pH and pKa,

pH=pK_a+log\frac{[A^-]_s}{[HA]_r}\; \; \; \; \; \; \; \; (2)

Eq2 is known as the Henderson-Hasselbalch equation, which is a good approximation of the pH curve after the start point and before the stoichiometric point for a strong base to weak acid titration with Ka ≤ 10-3 (or a strong acid to weak base titration with Kb ≤ 10-3).

 

Question

Rewrite eq2 to include the volume of acid Va and the volume of base Vb. What will the pH curve formula be if we disregard the 2nd and 3rd assumptions?

Answer

pH=pK_a+log\left ( \frac{[B]V_b}{V_a+V_b}/\frac{[A]V_a-[B]V_b}{V_a+V_b} \right )

pH=pK_a+log\left ( \frac{[B]V_b}{[A]V_a-[B]V_b} \right )\; \; \; \; \; \; \; \; (2a)

where [A] = [HA]ud , and [B] is the initial concentration of the monoprotic base. See this article for the derivation of the pH curve formula without the 2nd and 3rd assumptions.

 

The diagram below is the superimposition of a green curve (the Henderson-Hasselbalch equation, eq2a) on a blue curve (the complete pH titration curve, which disregards the above three assumptions), for the titration of 10 cm3 of 0.200 M of CH3COOH (Ka = 1.75 x 10-5) with 0.100 M of NaOH.

Even though the two curves appear to fit perfectly (apart from the initial bit before 1.00 cm3), they do not actually coalesce and are still two separate curves (discernible when the axes of the plot are scaled to a very high resolution). However, for practical purposes, the Henderson-Hasselbalch equation is a very good approximation of a pH curve for the region after the start point and before the stoichiometric point for a strong base to weak acid titration.

When [A–]s = [HA]r , eq2 becomes

pH=pK_a\; \; \; \; \; \; \; \; (3)

Eq3 occurs when the amount of salt formed equals to [HA]ud/2, i.e. when half of the acid is neutralised. This stoichiometric relation corresponds to the curve’s inflexion point, which is also called the maximum buffer capacity point.

 

Question

Why does the solution have maximum buffer capacity when half of the acid is neutralised, i.e. at the half-way volume point between the start point and the stoichiometric point?

Answer

For the buffer to be most effective, it has to resist to the greatest extent the change in pH versus the change in volume of base added. This means that the maximum buffer capacity corresponds to the point where the gradient of the function of eq2a is a minimum, i.e. at the inflexion point (see above diagram). 

Taking the 2nd derivative of eq2a with respect to Vb and letting \frac{d^2pH}{dV_b^2}=0,  we have

\frac{n_b}{n_a}=\frac{1}{2}

where n_b=\left [ B \right ]V_b  and n_a=\left [ A \right ]V_a.

Therefore, the inflexion point occurs when half of the total volume of base required for complete neutralisation of the acid is added.

If you want a full mathematical description of the maximum buffer capacity of a weak acidic buffer, read this article in the advanced section. 

 

For a strong acid to weak base titration, we have the following weak base equilibrium:

BOH(aq)\rightleftharpoons B^+(aq)+OH^-(aq)

K_b=\frac{[B^+][OH^-]}{[BOH]}

Taking the logarithm on both sides of the above equation and rearranging,

pOH=pK_b+log\frac{[B^+]}{[BOH]}\; \; \; \; \; \; \; \; (3a)

Eq3a is the basic form of the Henderson-Hasselbalch equation.

With reference to eq2 (or eq3a), knowing the pH of a strong base to weak acid titration (or knowing the pOH of a strong acid to weak base titration) at the half-way volume point, we can calculate the pKa (or pKb) of the weak acid (or weak base).

 

Question

Show that pKa + pKb = 14.

Answer

HA(aq)+H_2O(l)\rightleftharpoons A^-(aq)+H_3O^+(aq)

A^-(aq)+H_2O(l)\rightleftharpoons HA(aq)+OH^-(aq)

 

K_a=\frac{[A^-][H_3O^+]}{[HA]}\; \; \; \; \; \; \; \; K_b=\frac{[HA][OH^-]}{[A^-]}

pK_a+pK_b=-log\frac{[A^-][H_3O^+]}{[HA]}-log\frac{[HA][OH^-]}{[A^-]}

=-log[H_3O^+][OH^-]=-logK_w=14

 

 

next article: stoichiometric point of titration
Previous article: start point of titration
Content page of strong base to weak acid titration curve
Content page of intermediate chemistry
Main content page
Posted on

Classical atomic theory

The classical atomic theory is an early scientific interpretation of the atom.

Niels Bohr, a Danish physicist, introduced a model in 1911 to explain the hydrogen spectrum, which he used to derive the Rydberg formula.

He proposed that the electron of a hydrogen atom orbits around the fixed massive nucleus (see diagram below), with the coulombic force of interaction between the electron and the nucleus being balanced by the centrifugal force; that is:

\frac{e^2}{4\pi \varepsilon _0r^2}=\frac{m_ev^2}{r}\; \; \; \; \; \; \; \; 1

where e is the charge on the electron, ε0 is permittivity of free space, r is the radius of the orbit, me is the mass of the electron, and v is the speed of the electron.

Bohr further proposed that, for an orbit to remain stable, the angular momentum of the electron must be an integer multiple of \hbar=h/2\pi:

m_evr=n\frac{h}{2\pi}\; \; \; \; \; \; \; \; 2

where  h  is the Planck constant.

Combining eq1 and eq2 and eliminating v yields

r=\frac{\varepsilon _0n^2h^2}{\pi m_ee^2}\; \; \; \; \; \; \; \; 3

Substituting eq3 back in eq2 gives

v=\frac{e^2}{2\varepsilon _0nh}\; \; \; \; \; \; \; \; 4

 

Question

What is angular momentum and how did Bohr arrive at the assumption for eq2?

Answer 

Angular momentum is the rotational analogue of linear momentum. While the momentum of a body travelling in a straight line is proportional to its mass and velocity (p = mv),  the momentum of a body revolving around a point is proportional to its mass, tangential velocity, and the perpendicular distance from its tangential velocity to the point (L = mvr). According to Bohr, electrons orbit the nucleus at distinct radii and, consequently, have discrete values of angular momentum. He proposed that these radii are such that the angular momentum are integer multiples of h/2\pi.

Louis de Broglie, a French physicist, subsequently reinterpreted Bohr’s model by treating electrons as waves. He proposed that an integral number of the electron’s wavelength \lambda must fit the orbit’s circumference for the orbit to be stable, i.e. 2\pi r=n\lambda, where n\in \mathbb{Z}. Otherwise, the orbiting wave will disappear due to destructive interference (see diagram below).

Substituting de Broglie’s formula of p = h/λ into 2πr = nλ gives

p=\frac{nh}{2\pi r}\; \; \; \Rightarrow \; \; \; v=\frac{nh}{2\pi rm_e}\; \; \; and\; \; \; r=\frac{nh}{2\pi vm_e}\; \; \; \; \; \; \; (5)

Substituting eq5 in eq1 yields

v=\frac{e^2}{2\varepsilon _0nh}\; \; \; and\; \; \; r=\frac{\varepsilon _0 n^2h^2}{\pi m_ee^2}\; \; \; \; \; \; \; (6)

which is the same as eq4 and eq3, respectively.

 

The total energy of the electron is:

E_n=KE+V=\frac{1}{2}m_ev^2-\frac{e^2}{4\pi \varepsilon _0r}\; \; \; \; \; \; \; (7)

Substituting eq3 and eq4 in eq7 results in

E_n=-\frac{m_ee^4}{8{\varepsilon _{0}}^{2}n^2h^2}\; \; \; \; \; \; \; (8)

The transition energy between two states is:

\Delta E=E_2-E_1=\frac{m_ee^4}{8{\varepsilon _{0}}^{2}h^2}(\frac{1}{{n_{1}}^{2}}-\frac{1}{{n_{2}}^{2}})\; \; \; \; \; \; \; (9)

At this point, Bohr assumed that an allowed transition between two states involves an electron falling from a higher energy state to a lower energy state, with the emission of a photon of energy given by the Planck relation Ep = hv = ΔE. Therefore, eq9 becomes:

hv=hc\tilde{v}=\frac{m_ee^4}{8{\varepsilon _{0}}^{2}h^2}(\frac{1}{{n_{1}}^{2}}-\frac{1}{{n_{2}}^{2}})

\tilde{v}=\frac{m_ee^4}{8{\varepsilon _{0}}^{2}ch^3}(\frac{1}{{n_{1}}^{2}}-\frac{1}{{n_{2}}^{2}})\; \; \; \; \; \; \; (10)

where \tilde{v}=\frac{1}{\lambda} is the wavenumber of the photon.

When n1 = 1 and n2 = ∞,

\tilde{v}=\frac{m_ee^4}{8{\varepsilon _{0}}^{2}ch^3}=R_H\;\; or\; \; R_\infty \; \; \; \; \; \; \; (11)

where RH or R∞ is the Rydberg constant.

Even though the Bohr model has certain shortcomings – specifically, that an electron orbiting around the nucleus would constantly radiate electromagnetic energy and eventually crash into the nucleus – eq10 and eq11 were proven to be mathematically sound for the hydrogen atom using quantum mechanics.

From eq11,

m_e=\frac{8{\varepsilon _{0}}^{2}ch^3R_\infty }{e^4}=\frac{2hR_\infty }{c\alpha ^2}\; \; \; \; \; \; \; (12)

where \alpha =\frac{e^2}{2\varepsilon _0hc}\; \; or\; \; \frac{e^2}{4\pi \varepsilon_0\hbar c}\; \; \; \; \; \; \; \; (\hbar=\frac{h}{2\pi})

\alpha is known as the fine-structure constant.

 

Question

Show that the molar mass of carbon-12, M(12C), is:

M(^{12}C)=\frac{24hR_\infty N_A}{c\alpha ^2A_r(e)}

Answer 

\frac{molar\; mass\; of\; ^{12}C}{molar\; mass\; of\; electron}=\frac{relative\; mass\;of\; ^{12}C}{relative \; mass\; of\; electron}

Since the relative mass of 12C is 12 unified atomic mass units,

\frac{M(^{12}C)}{M(e)}=\frac{12}{A_r(e)}

From eq12, M(e)=\frac{2hR_\infty }{c\alpha ^2}N_A. Therefore,

M(^{12}C)=\frac{24hR_\infty N_A}{c\alpha ^2A_r(e)}

 

Next article: modern atomic structure
Content page of intermediate quantum chemistry
Content page of intermediate chemistry
Main content page
Posted on

Link between unified atomic mass and inertia mass

What is the link between unified atomic mass unit and inertia mass?

Jean Perrin’s and other scientists’ experiments in the early 1900s to determine the Avogadro constant were based on a gramme-molecule, which is the mass of a gas that occupies the same volume as two grammes of hydrogen gas at the same temperature and pressure. Their experiments produced a range of values for the constant when different gases are used. This variation occurs because different real gases with the same number of particles have different volumes.

A better definition of the mole was therefore needed and was established in 1967 as follows:

The amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12.

The choice to base the Avogadro constant on carbon-12 may seem arbitrary. However, it is this definition that enables us to link the unified atomic mass unit scale to the inertia mass scale. So,

1^{12}C=12u

\frac{0.012\; kgmol^{-1}}{N_{A}\; mol^{-1}}=12u\; \; \; \; \; \; \; (1)

1u=1.660539\times 10^{-27}\; kg\; \; \; \; \; \; \; (2)

We can rewrite eq1 as

1u=\frac{M_{u}}{N_{A}}g

where Mu is the molar mass constant and is equal to 1 gmol-1.

Even though the definition of the mole was changed to ‘a mole is the amount of substance of a system that contains exactly 6.02214076 x 1023 elementary entities’ in Nov 2018, the peg of 112C to 12 u remains. However, the exact value of the Avogadro constant leads to the molar mass of carbon-12 having a relative uncertainty in the order of 10-10. It is no longer exactly 0.012 kgmol-1 and is given by the formula (see the article on ‘Bohr model‘ for derivation):

M(^{12}C)=\frac{24hR_{\infty }N_{A}}{c\alpha^{2}A_{r}(e)}\; \; \; \; \; \; \; \; (3)

where

h   is the Planck constant

R_{\infty }   is the Rydberg constant

c   is the speed of light

\alpha   is the fine-structure constant

A_{r}(e)   is the ‘relative atomic mass’ of an electron

The uncertainty in the value of 0.012 kgmol-1, which will be determined in future experiments, is primarily due to the uncertainty in the fine-structure constant.

Similarly, the molar mass constant , which was a constant with the value 1 gmol-1, is now described by the formula:

M_{u}=\frac{2hR_{\infty}N_{A}}{c\alpha ^{2}A_{r}(e)}

By the same logic, the unified atomic mass unit has the formula:

u=\frac{2hR_{\infty }}{c\alpha^{2}A_{r}(e)}

 

next article: how to ‘weigh’ an atom?
previous article: unified atomic mass unit
Content page of mass in chemistry
Content page of intermediate chemistry
Main content page
Posted on

Millikan’s oil drop experiment: overview

Robert Millikan conducted an experiment in 1909 to measure the charge of an electron. He devised an apparatus to observe electrically-charged droplets of oil between two horizontal plates that had a potential difference of several thousand volts applied across them.

The oil droplets were sprayed from an atomiser into an upper chamber. In the space between the plates, the air were ionised by X-rays, causing the oil droplets to acquire electrons and become negatively charged as they fell and traversed through the air between the plates. Millikan then performed a few tasks and observed the motion of the oil droplets through a microscope.

 

Next ArTicle: Task 1
Content page of millikan’s oil drop experiment
Content page of intermediate chemistry
Main content page
Posted on

Millikan’s oil drop experiment: task 1

Millikan’s oil drop experiment consists of a few tasks.

For a start, the power supply to the plates is cut off (V = 0). The oil-droplets falling between the plates accelerate due to the effect of gravitational force and then continue downwards at terminal velocity as they encounter increasing air resistance. The three forces acting on an oil droplet are:

Gravitational force,

Fg

 The force of attraction between the mass of the earth and that of the oil droplet.
Upthrust (buoyant),

Fu

 The force exerted by a liquid or a gas on a body when the body displaces some weight of the liquid or gas. It is due to the pressure difference between the top and bottom of the object (as a result of the collision of liquid or gas molecules with the body from the top and bottom) and equal to the weight of the liquid or gas displaced.
Drag (viscous),

Fd

 The force caused by relative motion (friction) between an object and a liquid or gas. It is proportional to the velocity of the object, the radius of the object and the viscosity of the liquid or gas.

 

At terminal velocity,

F_{g}=F_{d}+F_{u}\; \; \; \; \; \; \; (1)

The oil droplet is assumed to be spherical. So, Fg, the weight of the oil droplet, is:

F_{g}=mg=\frac{4\pi }{3}r^{3}\rho g\; \; \; \; \; \; \; (2)

where m, ρ and r are the mass, density and radius of the oil droplet, respectively.

The equation of the drag force acting on the oil droplet can be expressed using Stokes’ law:

F_{d}=6\pi r\eta v_{1}\; \; \; \; \; \; \; (3)

where v1, is the terminal velocity of the falling oil droplet and η is the viscosity of air.

Through the microscope, Millikan recorded the distance travelled by the falling oil droplet over a period of time and calculated the terminal velocity v1.

For a perfectly spherical oil droplet the upthrust acting on it is:

F_{u}=mg=\frac{4\pi }{3}r^{3}\rho _{air}g\; \; \; \; \; \; \; (4)

where ρair is the density of the air.

Substituting eq2, eq3 and eq4 in eq1 yields

6\pi r\eta v_{1}=\frac{4\pi }{3}r^{3}(\rho -\rho _{air})g\; \; \; \; \; \; \; (5)

r=\sqrt{\frac{9\eta v_{1}}{2g(\rho -\rho _{air})}}\; \; \; \; \; \; \; (6)

NEXT ARTICLE: TASK 2
Previous article: Overview
Content page of millikan’s oil drop experiment
Content page of intermediate chemistry
Main content page
Posted on

Millikan’s oil drop experiment: task 2

The power supply to the plates is turned back on and the electric force acting on the oil droplet is:

F_{e}=qE\; \; \; \; \; \; \; (7)

where q is the charge on the oil droplet and E is the electric field between the plates.

For parallel plates,

E=\frac{V}{d}\; \; \; \; \; \; \; (8)

where V is the potential difference across the plates and d is the distance between the plates.

Combining eq7 and eq8,

q=\frac{F_{e}\: d}{V}\; \; \; \; \; \; \; (9)

If V is adjusted until the oil droplet remained stationary,

F_{e}=mg\; \; \; \; \; \; \; (10)

Combining eq9 and eq10,

q=\frac{mgd}{V}

Our objective is to determine the value of q, but it is difficult to accurately measure m. Therefore, we have to carry out another task.

 

NEXT ARTICLE: TASK 3
Previous article: task 1
Content page of millikan’s oil drop experiment
Content page of intermediate chemistry
Main content page
Posted on

Millikan’s oil drop experiment: task 3

Next, V is turned up slightly so that the oil droplet rises with a new terminal velocity v2.

We have:

F_{e}+F_{u}=F_{d}+F_{g}\; \; \; \; \; \; \; (11)

Note that the electric force is acting upwards (since the oil droplet is rising); upthrust is also acting upwards (because the pressure is still greater below the object than above it due to the exponential distribution of air); drag is now acting downwards because the oil droplet is rising, and weight continues to act downwards due to the effect of gravitational force. Substituting eq2, eq3 (with v2 instead of v1), eq4 and eq7 in eq 11 yields

qE+\frac{4\pi }{3}r^{3}\rho _{air}g=6\pi \eta rv_{2}+\frac{4\pi }{3}r^{3}\rho g

qE-\frac{4\pi }{3}r^3(\rho -\rho _{air})g=6\pi \eta rv_{2}\; \; \; \; \; \; \; (12)

Substituting eq5 in eq12 gives

qE-6\pi r\eta v_{1}=6\pi \eta rv_{2}\; \; \; \; \; \; \; (13)

Substituting eq8 in eq13 and rearranging results in

q=\frac{6\pi r\eta (v_{1}+v_{2})d}{V}\; \; \; \; \; \; \; (14)

Once again, Millikan recorded the distance travelled by the rising oil droplet over a period of time and calculated the new terminal velocity, v2.

Substituting the values of v1, v2 and the calculated value of r (from eq6) in eq14, Millikan determined a value for q. He repeated the experiment multiple times, each time varying the strength of X-rays ionizing the air, resulting in a varying number of electrons attaching to the oil droplet. He then obtained various values of q and found them to be multiples of 1.5924 x 10-19 C. Millikan concluded that the value of 1.5924 x 10-19 C is equal to the charge of a single electron. Subsequent determinations refined that value to 1.602176487 x 10−19 C.

Using the Faraday constant and the value of the charge of a single electron that he determined, Millikan calculated the Avogadro constant and proved that one Faraday is the quantity of charge for one mole of electrons.

 

Previous article: task 2
Content page of millikan’s oil drop experiment
Content page of intermediate chemistry
Main content page