Question

1) Consider the following acid ionization constants:

HA: Ka = 1.00*10-4	HB: Ka = 1.00*10-7
HC: Ka = 1.00*10-10	HD: Ka = 1.00*10-11

Solutions of each are prepared in which the inital concentration of the acid is 0.1000 M. Which of the four solutions will have the lowest pH?

the answer is HA

Explain why it is HA

2) A certain first order reaction is 30% complete in 115 minutes. what is the half life ( t1/2) of the reaction?

a) 192 min

b) 223 min

c) 268 min

e) 66.2 min

ANSWER ALL QUESTIONS AND EXPLAIN WHY YOU CHOSE THE ANSWER

Answer 1

1. Here all the acids are weak monobasic acids which are partially dissociated when dissolved in water.

Now consider the dissociation of HA. Let the degree of dissociation of HA be ₁

HA <-------------------> H⁺ + A^-

At eqilibrium C(1 - ₁) C₁ C₁

Where C = 0.1000M = concentration of the solution

Now equilibrium constant Ka for the above dissociation can be calculated as

Ka = [H+][A-]/[HA] =   (C₁)x(C₁)/C(1 - ₁)

Since for a weak acid  ₁ <<<< 1, hence 1 - ₁ nearly equals to 1.

Now Ka =  C₁²  

or ₁ = underroot(K_a/C)

Now [H+] can be calculated as

[H+] = C₁  = C x underroot(K_a/C) = underroot(K_a xC)

Hence higher the Ka value, higher is the [H+]

and P^H  = - log[H+] = - log underroot(K_a xC)

how ever due to negataive sign, higher the Ka value,lower is the P^H.

Since the Ka value of HA is the highest among all the given acids, it will have the lowest PH.

2. Let the initial amount of the reactant = A₀

Since the raction is 30 % completed in 115min, final amount of reactant, A = 70A₀/100 = 7A₀/10

t = 115 min

Rate constant for 1st order reaction can be calculated as

K = (1/t)x2.303log(A₀/A) = (1/115) x2.303logA₀/(7A₀/10) = (1/115) x2.303log(10/7)

=> K = 0.0031 min^-1

now half life can be calculated as

t_1/2 = 0.693/K = 0.693/0.0031 min^-1 = 223 min

Hence (b) is the correct answer.