Strong base is dissolved in 685 mL of 0.400 M weak acid (Ka = 3.88 x 10-5) to ma
ID: 957543 • Letter: S
Question
Strong base is dissolved in 685 mL of 0.400 M weak acid (Ka = 3.88 x 10-5) to make a buffer with a pH of 4.03. Assume that the volume remains constant when the base is added HA(aq) + OH-(aq) H2O(l) + A-(aq) lq, Calculate the pKa value of the acid and determine the number of moles of acid initially present. Number Number pKa = mol HA When the reaction is complete, what is the concentration ratio of conjugate base to acid? Number = [HA] HA How many moles of strong base were initially added? Number mol OHExplanation / Answer
pKa = -logKa = 4.411
Moles of acid present initially = molarity*volume of solution in litres = 0.4*0.685 = 0.274
Now, pH = pKa + log{[salt]/[acid]}
or, 4.03 = 4.411 + log{[A-]/[HA]}
or, [A-]/[HA] = 0.416
Thus, [A-] = 0.416*0.4 = 0.1664 M = [OH-]