Answer to Question #159556 in General Chemistry for Melissa

Question

1. What is the pH of a 0.350 M solution of lithium citrate? (K_a Citric Acid = 7.4 x 10^-4) 

2.Consider the following reaction:

PCl₃ (g)+ Cl₂ (g)   PCl₅ (g)                                      

K = 4.30 x 10^-6

Initially, 2.50 M PCl₃ and 1.40 M Cl₂ are placed into a container and allowed to reach equilibrium. Determine the equilibrium concentrations of all the species

Answer 1

1. PH = 8.33

2 . PCl₃ + Cl₂ $\to$ PCl₅

I = 2.50...1.40 .......0

C = -x ..... -x ........+x

E =(2.50-x).....(1.40-x).......x

K_c = $\frac{[PCl_5]}{[PCl_3][Cl_2]}$ = $\frac{x}{(2.50-x)(1.40-x)}$ = 4.30×10^-6

The equation

0.0000043x² - 1.00001677x + 0.00001505 = 0

has given 2 roots x = 0.0206 and x = 22,325.18 ( x = 22,325.18 which is not acceptable because its gives negative concentration for these) Now we taken x = 0.0206

[PCl₃] = 2.50 - x = 2.50 - 0.0206 = 2.294M

[Cl₂] = 1.40 - x = 1.40 - 0.0206 = 1.194M

[PCl₅] = x = 0.206M