Question

Predict whether or not a precipitate forms upon mixing 175.0 mLmL of a 0.0055 MKClMKCl solution with 145.0 mLmL of a 0.0015 MM AgNO3AgNO3 solution. Identify the precipitate, if any.

Answer 1

Answer : The precipitate is, AgCl.

Explanation :

The balanced chemical reaction will be:

`AgNO_3(aq)+KCl(aq)\rightarrow AgCl(s)+KNO_3(aq)`

First we have to calculate the molarity of
`Cl^-` ion.

`M_1V_1=M_2V_2`

where,

`M_1\text{ and }V_1` are the initial molarity and volume of KCl.

`M_2\text{ and }V_2` are the final molarity and volume of
`Cl^-` ion.

We are given:

`M_1=0.0055M\\V_1=175.0mL\\M_2=?\\V_2=175.0+145.0=320mL`

Putting values in above equation, we get:

`0.0055M* 175.0mL=M_2* 320mL\\\\M_2=0.00301M`

Now we have to calculate the molarity of
`Ag^+` ion.

`M_1V_1=M_2V_2`

where,

`M_1\text{ and }V_1` are the initial molarity and volume of
`AgNO_3`.

`M_2\text{ and }V_2` are the final molarity and volume of
`Ag^+` ion.

We are given:

`M_1=0.0015M\\V_1=145.0mL\\M_2=?\\V_2=175.0+145.0=320mL`

Putting values in above equation, we get:

`0.0015M* 145.0mL=M_2* 320mL\\\\M_2=0.000679M`

Now we have to calculate the value of reaction quotient.

The expression for reaction quotient will be :

`Q_(sp)=[Ag^+][Cl^-]`

Now put all the given values in this expression, we get

`Q_(sp)=(0.000679)* (0.00301)=2.04* 10^(-6)`

The given solubility constant value is,
`K_(sp)=1.77* 10^(-10)`

• When
  `K_(sp)>Q_(sp)`; the reaction is product favored. (No precipitation)
• When
  `K_(sp)<Q_(sp)`; the reaction is reactant favored. (Precipitation)
• When
  `K_(sp)=Q_(sp)`; the reaction is in equilibrium. (Sparingly soluble)

As,
`K_(sp)<Q_(sp)` then the reaction is reactant favored that means formation of precipitation.

Thus, the precipitate is, AgCl.