A precipitate is expected when an aqueous solution of porassium iodide is added to an aqueous solution of
AgNO3. Actually there are several answers that are correct. Ag^+, Hg2^+2, Hg+2, Pb^+2 will form the corresponding Iodide that will be insoluble in water.
would lead nitrate make a soluble
PbI2 is insoluble in water, as is AgI, Hg2I2, HgI2.
To determine whether a precipitate will form when an aqueous solution of potassium iodide (KI) is added to another aqueous solution, such as silver nitrate (AgNO3), you need to consider the solubility rules.
1. Start by identifying the ions present in each solution. In this case, the potassium iodide solution contains K+ and I- ions, while the silver nitrate solution contains Ag+ and NO3- ions.
2. Next, consult a solubility table or the solubility rules to determine if the combination of these ions will form a precipitate. The solubility rules are guidelines that indicate whether an ionic compound will remain dissolved in water or form a solid precipitate.
3. According to the solubility rules, most salts containing the nitrate (NO3-) ion are soluble and do not form precipitates. Therefore, the nitrate ion (NO3-) from silver nitrate will remain dissolved.
4. Iodide (I-) is also generally soluble, but it forms an exception when combined with silver (Ag+). According to the solubility rules, silver iodide (AgI) is insoluble and will form a yellow precipitate.
Therefore, when you add an aqueous solution of potassium iodide (KI) to an aqueous solution of silver nitrate (AgNO3), a yellow precipitate of silver iodide (AgI) will form. This reaction can be represented by the following equation:
KI(aq) + AgNO3(aq) -> AgI(s) + KNO3(aq)
Remember, solubility rules are essential in predicting whether a precipitate will form, but there can be exceptions. Always consult reliable references or experimentally determine solubilities when necessary.