Strong base is dissolved in 615 mL of 0.200 M weak acid (Ka = 3.24 × 10-5) to ma
ID: 954036 • Letter: S
Question
Strong base is dissolved in 615 mL of 0.200 M weak acid (Ka = 3.24 × 10-5) to make a buffer with a pH of 4.07. Assume that the volume remains constant when the base is added.
Calculate the pKa value of the acid and determine the number of moles of acid initially present.
When the reaction is complete, what is the concentration ratio of conjugate base to acid?
How many moles of strong base were initially added?
Strong base is dissolved in 615 mL of 0.200 M weak acid (Ka 3.24 x 10) to make a buffer with a pH of 4.07 Assume that the volume remains constant when the base is added. HA(aq)-OH-(aq) H,09-A-(aq) Calculate the pKa value of the acid and determine the number of moles of acid initially present. Number Number pK mol HA When the reaction is complete, what is the concentration ratio of conjugate base to acid? Number HA How many moles of strong base were initially added? Number mol OHExplanation / Answer
Given Ka = 3.24 x 10-5
=> pKa = - logKa = - log(3.24 x 10-5 ) = 4.49 (answer)
For the weak acid, concentration, M = 0.200 M
Volume, V = 615 mL = 0.615 L
Hence moles of the acid initially present = MxV = 0.200 mol/L x 0.615 L = 0.123 mol (answer)
Given the pH of the buffer solution = 4.07
Applying Hendersen equation
pH = pKa + log [A-] / [HA]
=> 4.07 = 4.49 + log [A-] / [HA]
=> [A-] / [HA] = 10-0.42 = 0.380 (answer)
The neutralization reaction is
--------------- HA(aq) + OH-(aq) ------- > A-(aq) + H2O(l)
Initial mol: 0.123, --- y mol -------------- 0
---change: - y mol, - y mol ------------ + y mol
Final.mol: (0.123 - y), 0 mol, ---------- y mol
=> y mol / (0.123 - y) mol = 0.380
=> 1.380y = 0.123 x 0.380
=> y = 0.0339 mol
Hence moles of strong base initially added = 0.0339 mol (answer)