Question

1) Worked these out and I just want to make sure I got them right at each step.

a) Given: A yields C + D K = 5.0 x 10^-8 with

[A] initial = 15*10^-5

[C] initial = 20 * 10^-10

[D] initial = 10 *10^-4

Determine the equilibrium concentrations of each:

In this one I believe you solve for the reaction quotient Q, first.

Equation: Q = Products/Reactants or Q= [C] [D] / [A] (use initial concentrations) Then from here you determine if Q < K or Q > K. This tells you which way the equilibrium will shift and if the pH will be less than or greater than 7 at equilibrium. This values also helps you to know which signs to use for the x when solving for the equilibrium concentrations (this part I'm not sure about. I'm not sure if I used the right sign or if I used the right value for x (if there should be something subtracted from one of them?).

One side of the reactions is supposed to be negative, while the other side is positive. Just not sure how to get that.

A yields C + D

I (initial concentrations) 15* 10^-5 20*10^-10 10*10^-4 C

-x + x^2 + x

E (equilibrium concentrations)

Then you use the equation for K, which is K= Products/Reactants, substituting in the values of x for the concentrations (don't use just the initial concentrations). From this you find x, which can then be used to find the equilibrium concentrations.

b) Given a 50 mL buffer solution: K_[HA] = 10^3 [HA] = 1.0 M [A^-] = 1.0 M Determine the pH with the addition of 10 mL of a 2 M HCl solution:

I think in this one you use the Henderson-Hasselbalch equation, pH = pKa + log [A-]/[HA].

I'm not sure what to do so I just included all equations related to buffers and Acid-Base equilibria.

*weak acid (pKa = -log Ka)

pH = -log [H3O-]

pOH = -log [OH-] [H3O+] = 10^-pH

[OH-] = 10^-pOH

Ka = [H3O+] [A-] / [HA]

pKa = -log Ka

Answer 1

#1(a): Yes, first we have to find the value of Q.

Here Q = 1.33*10^-8 which is less than K i.e Q <K.

Hence the equilibrium will shift to right side.

----A --------------> C + D ; K = 1.33*10^-8

I: 15*10^-5 ----- 20*10^-10, 10*10^-4

C: - X ------------ +X ------ + X

E: (15*10^-5-X), (20*10^-10+X), (10*10^-4+X)

K = 1.33*10^-8 = [(20*10^-10+X)*(10*10^-4+X)] / (15*10^-5-X)

By solving the above equation, we can find the value of X and then find the equilibrium concentrations as:

[A] = (15*10^-5-X)

[C] = (20*10^-10+X)

[D] = (10*10^-4+X)  

#1(b): To solve this question, first we have to write the balanced neutralization equation and then ICE chart to find the moles of HA and A^- remained after the reaction is over. Also we have to find the moles of HCl, HA ans A-.

Moles of HCl added = C*V = 2 mol/L * 0.010 L = 0.020 mol

Moles of HA added = C*V = 1 mol/L * 0.050 L = 0.050 mol

Moles of A^- added = C*V = 1 mol/L * 0.050 L = 0.050 mol

H⁺ of HCl rects with A^- to form HA. Hence the concentration HA increases and that of A- decreases.

--- H⁺ (aq) + A-(aq) -------> HA

I: 0.020 mol, 0.050 mol, 0.050 mol

C: - 0.020 mol,- 0.020 mol, + 0.020 mol

E: 0 mol, ----- 0.030 mol, ---- 0.070 ml

Now use Henderson equation:

pH = pKa + log[A-(aq)] / [HA(aq)]

Sice volume is same for all, we can replace concentration by mole

=> pH = pKa + log(0.030 / 0.070) (Answer)