Question

Calculate the number of milliliters of 0.753 M KOH required to precipitate all of the Pb2+ ions in 135 mL of 0.775 M Pb(NO3)2 solution as Pb(OH)2. The equation for the reaction is: Pb(NO3)2(aq) + 2 KOH(aq) Pb(OH)2(s) + 2 KNO3(aq) mL KOH

Answer 1

Answer:

278 ml of KOH.

Explanation:

The equation for the reaction is: Pb(NO3)2(aq) + 2KOH(aq) --> Pb(OH)2(s) + 2KNO3(aq)

Pb(NO3)2:

Half ionic equation:

Pb(NO3)2(aq) --> Pb2+ + 2NO3^-

Pb(OH)2 --> Pb2+ + 2OH-

Volume = 135 mL

Molar concentration = 0.775 M

Number of moles = molar concentration * volume

= 0.775 * 0.135

= 0.105 mol of Pb(NO3)2

Since 1 mole of Pb(NO3)2 reacted with 2 moles of KOH to give 1 mole of Pb2+ (Pb(OH)2).

By stoichiometry,

Number of moles of KOH = 0.105 * 2

= 0.21 mol

Molar concentration = number of moles/volume

Volume = 0.21/0.753

= 0.278 l

To ml, 0.278 l * 1000 ml/1l

= 278 ml of KOH.

Answer 2

The number of milliliters of 0.753 KOH that will precipitate the PB2+ ions would be 278 mL

From the equation of the reaction:

$Pb(NO_3)_2(aq) + 2 KOH(aq) ---> Pb(OH)_2(s) + 2 KNO_3(aq)$

The mole ratio of KOH to Pb(NO3)2 is 2:1.

Mole of Pb(NO3)2 = molarity x volume

                             = 0.775 x 135/1000

                              = 0.105 moles

Thus, the mole of KOH would be:  0.105 x 2

                                               = 0.21 moles

Hence, volume of KOH = mole/molarity

                         = 0.21/0.753

                                = 0.278 L

                                = 278 mL

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