Answer:
the answer is not correct
Answer:
K^+ and NO3^-
Explanation:
In a balanced ionic equation, we usually see the species that react to yield the main product in the reaction.
Consider the reaction;
Pb(NO3)2(aq) +2 KI(aq) -------> PbI2(s) + 2KNO3(aq)
The main product in this reaction is PbI2. Hence the balanced ionic equation is;
Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)
Notice that K^+ and NO3^- did not participate in this reaction. All ions that are part of the molecular equation but do not participate in the ionic reaction equation are called spectator ions. Hence K^+ and NO3^- are spectator ions in this reaction as can be seen clearly above.
The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
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