Part A) Find the percent dissociation of a 0.140 MM solution of a weak monoprotic acid having Ka=2.0×10−3

Sure, here is the step-by-step solution:

Step 1: Write the dissociation equation for the weak monoprotic acid (HA). HA ⇌ H+ + A-

Step 2: Write the expression for the acid dissociation constant (Ka). Ka = [H+][A-]/[HA]

Step 3: Set up an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations of the species involved in the reaction.

Initial: [HA] = 0.140 M, [H+] = 0, [A-] = 0 Change: [HA] = -x, [H+] = +x, [A-] = +x Equilibrium: [HA] = 0.140 - x, [H+] = x, [A-] = x

Step 4: Substitute the equilibrium concentrations into the Ka expression and solve for x. 2.0×10−3 = x^2/(0.140 - x)

Assuming x is much smaller than 0.140, the equation simplifies to: 2.0×10−3 = x^2/0.140

Solving for x gives the concentration of H+ ions, which is approximately 0.0053 M.

Step 5: Calculate the percent dissociation using the formula: % dissociation = ([H+]/[HA]) * 100%

Substituting the values gives: % dissociation = (0.0053/0.140) * 100% = 3.79%

So, the percent dissociation of the weak monoprotic acid is approximately 3.79%.