Question

1. What is the value of Q (reaction quotient) for the solubility of BaCO3 when 249 mL of 0.075 M solution of Na2CO3(aq) are mixed with 207 mL of 0.072 M solution of BaCl2(aq).

2. To a solution contaminated with lead enough NaCl is added to bring the concentration of chloride ion up to 0.538M. What concentration of lead will remain in the solution? In other words, what is the solubility of lead (II) chloride in this solution? for PbCl₂, K_sp = 1.5 x 10^-5

^3.What is the molar mass of the acid if a titration of 4.56 g of an acid requires 43.01mL of 0.250M NaOH to reach the equivalence point?

Answer 1

(1)

Moles of Ba²⁺ = Molarity*Volume = 0.072*0.207 = 0.014904

Moles of CO₃^2- = 0.249*0.075 = 0.018675

Final volume = 249+207 = 456 mL = 0.456 L

So,

Q = [Ba²⁺]*[CO₃^2-] = (0.014904/0.456)*(0.018675/0.456) = 1.34*10^-3

(2)

Using relation:

[Pb²⁺][Cl^-]² = 1.5*10^-5

Putting values and solving we get:

[Pb²⁺]*(0.538)² = 1.5*10^-5.

Thus,

[Pb²⁺] = 5.18*10^-5 M

(3)

Moles of base required = Molarity*Volume = 0.250*0.04301 = 0.0107525

Assuming that the acid is monoprotic, the MW of acid is:

MW = Mass/moles = 4.56/0.0107525 = 424.09 g

Hope this helps !