Strong base is dissolved in 695 mL of 0.400 M weak acid (Ka = 3.77 × 10-5) to ma

Question

Strong base is dissolved in 695 mL of 0.400 M weak acid (Ka = 3.77 × 10-5) to make a buffer with a pH of 3.94. Assume that the volume remains constant when the base is added.

Strong base is dissolved in 695 mL of 0.400 M weak acid (Ka 3.77 × 10°) to make a buffer with a pH of 3.94. Assume that the volume remains constant when the base is added. HA(aq) + OH . (aq) H,04-A-(aq) Calculate the pKa value of the acid and determine the number of moles of acid initially present. Number Number mol HA When the reaction is complete, what is the concentration ratio of conjugate base to acid? A- [HA]-10 Number How many moles of strong base were initially added? Number mol OH

Explanation / Answer

1) we know that

pKa = -log Ka

so

pKa = -log 3.77 x 10-5

pKa = 4.42366

so

pKa of the acid is 4.42366

2)

now

we know that

moles = molarity x volume (L)

so

moles of acid = 0.4 x 695 x 10-3

moles of acid = 0.278

3)

we know that

for buffers

pH = pKa + log [A-/HA]

soo

3.94 = 4.42366 + log [A-/HA]

[A-/HA] = 0.32835

4)

let y moles of OH- be added

the reaction is

HA + OH- --> A- + H20

then

[HA] = 0.278 - y

[A-] = y

so

y / (0.278-y) = 0.32835

y = 0.0687

so

0.0687 mol of OH- is added

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