The quantitative relations for the chloride are as follows: a liter of water dissolves 0.002 gram, or 1.4E−5 mole, of silver chloride at 25°,[435] and the solubility-product constant at 25° is [Ag^{+}] × [Cl^{−}] = (1.4E−5)^2 = 2E−10. Now, if 1 c.c. of 0.1 molar sodium chloride is added to 10 c.c. of 0.05 molar silver-ammonium nitrate, we have,[436] for the first moment, [Ag^{+}] = 8.9E−4 and [p222] [Cl^{−}] = 0.008, and [Ag^{+}] × [Cl^{−}] = 8.9E−4 × 0.008 = 7E−6, which is much larger than the solubility-product constant, and precipitation must take place. The precipitate will be quite a heavy one: as silver-ion is removed from solution, the complex ion must decompose and furnish a new supply of silver-ion, and precipitation must continue until the excess of ammonia, which is liberated by the decomposition of the complex ion (Ag(NH_{3})_{2}^{+} + Cl^{−} → AgCl ↓ + 2 NH_{3}), suppresses the silver-ion sufficiently to satisfy, with the diminished concentration of chloride-ion, the solubility-product constant of silver chloride. It is clear that, while the reactions of silver-ion are not obtained ‹as readily› in the ammoniacal solution as in an equivalent solution of silver nitrate (bromate experiment), nevertheless more sensitive tests show that a ‹small portion› of the silver still is present as ‹silver-ion› in the ammoniacal solution (chloride experiment). This brings us to our third point, the influence of an excess of ammonia on the concentration of silver-ion and on its reactions. It is evident, from the form of the equilibrium equation (p. 219), that any excess of ammonia must very rapidly reduce the concentration of silver-ion. We may ask ‹what excess will be required to prevent the precipitation of silver chloride› in the experiment just tried. The question may be answered as follows: The concentration of chloride-ion, when 1 c.c. of 0.1 molar sodium chloride is added to 10 c.c. of 0.05 molar [Ag(NH_{3})_{2}]NO_{3}, no precipitate being formed, will be 0.1 × (1 / 11) × 0.87, the solution being diluted 1 to 11 and the percentage of ionization of a salt MeX being approximately 87% in 0.05 to 0.06 molar concentration. For a solution containing this concentration of chloride-ion, the concentration [Ag^{+}] of silver-ion, ‹which may just be present› «without» ‹leading to the precipitation of silver chloride› (‹i.e.› for the saturated solution) is, according to the principle of the solubility-product, [Ag^{+}] = K_{S.P.} / [Cl^{−}] = (2E−10) / (0.1 × 0.87 × 1 / 11). Further, in the presence of an excess of ammonia, practically all of the silver is present in the complex form, and, [Ag(NH_{3})_{2}^{+}] = 0.05 × 0.87 × 10 / 11 the 0.05 molar solution being diluted 10 parts to 11 and the salt being 87% ionized. If we call ‹x› the concentration of free ammonia required to reduce the concentration of silver-ion to the small value indicated, we may put