The Rate of Photographic Development

Background

Introduction to the Chemistry of Photography

Figure 1 shows the steps involved in producing a black and white photograph of an object, a white "T" on a black background in this case. The image of the object is focused on to the photographic film in the camera (Figure 1a). When the film is developed, a black and white negative results. The negative is black where light struck the film, and clear where no light struck. The steps involved in converting the negative to a "positive" or a photograph, are shown in Figure 1b. The negative is placed in contact with photographic paper and light is passed through the negative. Light strikes the photographic paper where the negative is clear and no light strikes the paper where the negative is black. If you examine the photographic paper after it has been exposed to light, the paper will appear to be unchanged. However, if the paper is placed in a photographic developer, a positive image of the original object will appear on the paper. Note that the finished photograph is black where light struck the paper and white where no light struck. The effect of light is to blacken the photographic paper (or photographic film).

The Emulsion

The emulsion is the light-sensitive coating on photographic film or photographic paper. Basically, it consists of microscopic crystals of silver halide (AgCl, AgBr or mixtures of AgBr and AgI) dispersed in gelatin. The crystals are a few microns in diameter. (1 micron = 10^-6 meters, one millionth of a meter.) There are about one billion (10⁹) crystals in each square centimeter of emulsion. Within each crystal of AgBr the ions Ag⁺¹ and Br^-1 are arranged in an orderly fashion as shown in Figure 2. In this figure the spheres represent the individual ions and the lines help to visualize the three-dimensional structural of the crystal. Remember this represents only a small part of a crystal. A real crystal will extend in all directions to include millions of ions.

The Latent Image

As indicated in Figure 1 b, no image is evident on the photographic paper until the paper is developed. How does the emulsion "remember" where it was struck by light? The answer follows below.

Light energy causes an electron, e^-, to be removed from a bromide ion, Br^-, in an AgBr crystal,

light energy + Br^{- ____}> Br + e^-

The electron becomes trapped in a region of the crystal called a sensitivity speck, SS

SS + e^{- ____}> SS ^-

The free Ag⁺ ions which can move around within the crystal are attracted to the negatively charged sensitivity specks and are converted to neutral silver atoms:

Ag⁺ + SS ^{- ____}> SSAg⁰

The process is repeated as more light hits the emulsion:

e^- + SSAg^{0 ____}> SS Ag^-

SS Ag^- + Ag^{+ ____}> SS Ag₂

and continues:

SS Ag^-_n-1 + Ag^{+ ____}> SS Ag_n

In the last reaction, SSAg_n represents a cluster of neutral silver atoms containing n atoms where n>4. Therefore, all of the AgBr crystals which have been struck by light contain tiny invisible clusters of neutral silver atoms. These tiny specks make up the "latent" or invisible image in the emulsion. Crystals which have not been struck by light do not contain these clusters. The latent image is the mechanism by which the paper "remembers" where it was struck by light.

When the emulsion is developed, only those crystals containing the clusters of metallic silver will turn black. the others remain colorless. The blackening of the crystals containing the latent image clusters is due to the reduction of the silver ions to silver metal by the developer. That is,

developer + Ag⁺ (in crystal containing silver atom clusters) ---> Ag⁰

developer + Ag⁺ (in crystal without silver atom clusters) --> no reaction

This is summarized below.

The conversion of the latent image to a stable finished photograph involves placing the photographic paper successively into each of four different solutions: the developer solution, the stop bath, and the fixer solution. The chemistry of each one of these steps is described below.

The Developer

The developer used in this experiment contains the following components:

1) Metol, ---the reducing agent

2) Sodium carbonate, (Na₂CO₃•H₂O) ---the activator

3) Sodium bromide, (NaBr) ---the restrainer

4) Sodium sulfate (Na₂SO₃) ---the preservative

5) Water, H₂O ---the solvent

The reducing agent and activator, Metol and Sodium Carbonate -- The reducing agent, Metol, in this case, reacts with the silver ions in the silver halide crystals which have been struck by light (that is, contain clusters of neutral silver atoms) and converts them to silver metal. In order for this reaction to occur, the solution must be basic in order to activate the Metol.

In other words, it is the "active" form of metol which actually reacts with the silver ions. The purpose of the activator, Na₂CO₃•H₂O, is to make the solution basic, so that the activated form of metol is produced by the first reaction above, When Na₂CO₃ is dissolved in water, the carbonate ion (CO₃^-2) which is formed reacts with the water to form hydroxide ion (OH^-) making the solution basic and reacting with the Metol to activate it.

Na₂CO₃•H₂O + H₂O ---> 2Na⁺ + HCO₃^- + OH^- + H₂O

The restrainer, NaBr -- The purpose of the restrainer is to prevent "fogging", the development or blackening of silver halide crystals which have not been struck by light.

The preservative, Na₂SO₃ -- The preservative serves to preserve the developer from the action of oxygen. The presence of the preservative slows down the spoilage of the developer, just as food preservatives slow down the spoilage of food.

The solvent, H₂O -- In order for the various components of the developer to do their job, they must get into the emulsion and make contact with the crystals of AgX. the developer components are all soluble in water, and the water carries them into the emulsion. Once the development is complete, water is used to wash the developer components out of the emulsion.

The Stop Bath

The purpose of the stop bath is to stop development of the photographic paper. If development is not stopped, the entire piece of photographic paper will turn black, and no print will be obtained. The stop bath is an aqueous solution of acetic acid, CH₃CO₂H, which reacts in water to give H⁺ ion

CH₃CO₂H ---> CH₃CO₂^-1 + H⁺¹

In an acidic solution the active form of metol is converted back to the inactive form, and development stops,

The Fixer

Once the development process is stopped, the emulsion still contains crystals of AgX which have not been struck by light. These are light sensitive, and must be removed. Otherwise, they will eventually turn black when they are exposed to white light. If this happens, the entire image will turn black. The AgX crystals are insoluble in water, and because of this, they cannot be washed away with pure water. The fixer solution contains sodium thiosulfate, Na₂S₂O₃, which dissolves in water to give thiosulfate ions, S₂O₃^2-:

Na₂S₂O₃ ^____> 2Na⁺(aq) + S₂O₃^2-(aq)

The S₂O₃^2- ions react with AgX to make a soluble compound, Ag(S₂O₃)₂^3-:

AgX + S₂O₃^{2- ____}> Ag(S₂O₃)₂^3- + Br^-

Thus, the AgX dissolves and can be washed away. The paper is no longer light sensitive.

The Wash

The final step involves washing the print in water to remove all of the chemicals used in the previous steps, some of which would render the print unstable if allowed to remain.

comments to: j byrd jim@chem.csustan.edu or m perona mike@chem.csustan.edu

02.03.99