Question

A buffer is made by adding 0.300 mol of CH3COOH and 0.300 mol of CH3COONa to enough water to make 1.000 L of solution. The pH of the buffer is 4.74.

(a) Calculate the pH of this solution after 5.00 mL of 4.0 M NaOH (aq) solution is added. Write out the balanced equation for the reaction.

(b) For a comparison, calculate the pH of a solution made by adding 5.0 mL of 4.0 M NaOH (aq) solution to 1.000 L of water.

(c) Did your calculated pH’s match what you expected? Please explain for either answer.

Answer 1

mmol of acid initially = MV! = 0.3*10^3 = 300

mmol of conjugate initially = MV = 0.3*10^3 = 300

a)

mmol of base = MV = 5*4 = 20 mmol of OH-

after reaction

mmol of acid left = 300-20 = 280

mmol of conjugate formed = 300+20 = 320

apply buffer equation, which is henderson haselbach equation

pH = pKa + log(A-/HA)

substitute known data

pH = 4.75 + log(320/280)

pH = 4.8079

The reaction (net ionic)

H+ + OH---> H2O

b)

pH of basic solutino

[OH-] = mol/V = (5*4)/(1000) = 0.2 M

pOH = -log(OH) = -log(0.2 = 0.698970

pH = 14-pOH = 14-0.698970

pH = 13.30

c)

Yes, the pH change is due to the buffer peresence A- and HA