Complete the balanced overall ionic equation for sodium iodide dissolving in wat

Question

Complete the balanced overall ionic equation for sodium iodide dissolving in water. Nal(s) ? Part 2 (1 point) 0 See Hint Complete the balanced overall ionic equation for lead(II) nitrate dissolving in water. Pb(NO3)2(s) ? Part 3 (1 point) 0 See Hint What will happen if we combine the solution of sodium iodide and the solution of lead(ll) nitrate? The solutions will mix, so all of the ions will be evenly distributed throughout the entire solution. But sometimes mixing solutions containing dissolved ions leads to a precipitation reaction: the formation of a new insoluble compound. To determine if this happens, we refer to the solubility rules for common ionic compounds in water (Table 4.4, p. 170) Based on the information presented in that table, complete the balanced overall ionic equation for the mixing of these two solutions. 2Na"(aq) + 21-(aq) + Pb2+ (aq) + 2NO,(ag)-? ?10:15 > 10OF 15 QUESTIONS COMPLETED

Explanation / Answer

a)

sodium iodide in water will form aqueous Na+ ions and I- in solution

NaI(s) = Na+(aq) + I-(aq)

the ionic equation must include all ions

NaI(s) = Na+(aq) + I-(aq)

b)

simlar to the salt above, Pb(NO3)2 will split in ions

then

Pb(NO3)2(s) --> Pb+2(aq) + 2NO-(aq)

the ionic equation must include both species

c)

overall ionic equation must include the solid formation, sinc ePb+2 + 2I- will not be soluble

2Na+(aq) + 2I-(aq) + Pb+2(aq) + 2NO3-(aq) = PbI2(s) + 2Na+(aq) + 2NO3-(aq)

the net ionic:

2I-(aq) + Pb+2(aq) = PbI2(s)

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