Question

In the laboratory 6.67 g of Sr(NO3)2 is dissolved in enough water to form 0.750 L. A 0.100 L sample is withdrawn from this stock solution and titrated with a 0.0460 M solution of Na3PO4. a. What is the concentration of the Sr(NO3)2stock solution? b. Write a balanced molecular equation for the titration reaction. c. How many milliliters of the Na3PO4 solution are required to precipitate all the Sr2+ ions in the 0.100 L sample? (MM's: Sr(NO3)2 = 211.64; Na3PO4 =163.94)Name

Answer 1

The answer to your question is below

Step-by-step explanation:

Data

mass of Sr(NO₃)₂ = 6.67 g Final volume = 0.750 L

Sample 0.100 L

[Na₃PO₄] = 0.046 M

a) [Sr(NO₃)₂

MM = 211.64 g

211.64 g ---------------------- 1 mol

6.67 g ---------------------- x

x = (6.67 x 1) / 211.64

x = 0.032 mol

Molarity = 0.032 / 0.75

Molarity = 0.042

b)

3Sr(NO₃)₂ + 2Na₃PO₄ ⇒ Sr₃(PO₄)₂ + 6NaNO₃

Reactants Elements Products

3 Sr 3

6 N 6

6 Na 6

2 P 2

24 O 24

c)

Calculate the moles of Sr(NO₃)₂ in 100 ml or 0.1 L

Molarity = moles / volume

Moles = Molarity x volume

Moles = 0.042 x 0.1

Moles = 0.0042

3 moles of Sr(NO₃)₂ --------------- 2 moles of Na₃PO₄

0.0042 moles of Sr(NO₃)₂ -------- x

x = (0.0042 x 2) / 3

x = 0.0028 moles of Na₃PO₄

Molarity = moles / volume

Volume = moles / Molarity

Volume = 0.0028 / 0.046

Volume = 0.060 L or 60.9 mL