Consider the dissolution equation of lead(II) chloride.
PbCl2 (s) ⇆ Pb2+ (aq) + 2Cl− (aq)
Suppose you add 0.2307 g of PbCl2 (s) to 50.0 mL of water. In the resulting saturated solution, you find that the concentration of Pb2+ (aq) is 0.0159 M and the concentration of Cl− (aq) is 0.0318 M.
What is the value of the equilibrium constant, Ksp, for the dissolution of PbCl2?

The solubility product of lead(II) chloride is .

Explanation:

Concentration of lead (II) ions =

Concentration of chloride ion =

The expression of a solubility product will be given as:

The solubility product of lead(II) chloride is .

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol