Question

What volume of 0.100 M Na3PO4 is required to precipitate all the lead(ii) ions from 150.0 mL of 0.250 M Pb(NO3)2?

Answer 1

First, we write the reaction equation:

3Pb(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + 3Pb₃(PO₄)₂

Moles of Pb ions present:
moles = concentration x volume
= 0.15 x 0.25
= 0.0375

From the equation,
moles Pb : moles Na₃PO₄
= 3 : 2
Moles of Na₃PO₄:
2/3 x 0.0375
= 0.025

volume = moles / concentration
= 0.025 / 0.1
= 0.25 L
= 250 ml

Answer 2

Volume of 0.100 M Na₃PO₄ is 0.25 L

Further explanation

Stoichiometry in Chemistry learn about chemicals mainly emphasizes quantitative, such as the calculation of volume, mass, number, which is related to numbers, molecules, elements, etc.

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

In the reaction there are also manifestations of reagent substances namely gas (g), liquid (liquid / l), solid (solid / s) and solution (aqueous / aq).

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

• Mole

The mole itself is the number of particles contained in a substance amounting to 6.02.10²³

Mole can also be sought if the amount of substance mass and its molar mass is known

$\large{\boxed{\boxed{\bold{mol=\frac{mass}{molar\:mass}}}}$

Reaction that happens :

2Na₃PO₄ + 3Pb(NO₃)₂ ⇒ 6NaNO₃ + 3Pb₃(PO₄)₂

Pb(NO₃)₂ ⇒ Pb²⁺ + 2NO₃⁻

mole Pb²⁺  = mole Pb(NO₃)₂

mole Pb²⁺ = 0.15 L(150 ml) . 0.25 M = 0.0375 mole

mole Na₃PO₄ : mole Pb(NO₃)₂ = 2:3

mole Na₃PO₄ = 2/3 . 0.0375

mole Na₃PO₄ = 0.025

So volume of 0.100 M Na₃PO₄ =

volume = mole : M

volume = 0.025 : 0.1 = 0.25 L

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Keywords: mole,  precipitate, Na₃PO₄, Pb(NO₃)₂, volume