Strong base is dissolved in 715 mL of 0.400 M weak acid (Ka = 4.38 × 10-5) to ma

Question

Strong base is dissolved in 715 mL of 0.400 M weak acid (Ka = 4.38 × 10-5) to make a buffer with a pH of 4.01. Assume that the volume remains constant when the base is added. HA(aq) + OH- (aq) -->H2O(l) + A-(aq)
A.) Strong base is dissolved in 715 mL of 0.400 M weak acid (Ka = 4.38 × 10-5) to make a buffer with a pH of 4.01. Assume that the volume remains constant when the base is added. HA(aq) + OH- (aq) -->H2O(l) + A-(aq)
A. )I found the PKA, which is 4.36 and the mol of HA is .286.
B.)When the reaction is complete, what is the concentration ratio of conjugate base to acid? [A-]/[HA]=?
C.)How many moles of strong base were initially added? moL[OH-]= ?

Explanation / Answer

we know that

for buffers

pH = pKa + log [ conjugate base / acid ]

so

pH = pKa + log [ A- / HA]

given

pH = 4.01

pKa = 4.36

so

4.01 = 4.36 + log [ A- / HA]

[A-/HA] = 0.447

the ratio is 0.447

c)

we know that

moles = molarity x volume (L)

so

initially

moles of HA = 0.4 x 0.715 = 0.286

let

moles of base added be y

now

the reaction is

HA + OH- --> A- + H20

now

moles of HA reacted = moles of OH- added = y

moles of A- formed = moles of HA reacted = y

now

finally

moles of HA = 0.286 - y

moles of A- = y

now

[A- / HA] = 0.447

so

y / ( 0.286 -y) = 0.447

y = 0.089

so

0.089 moles of strong base were initially added

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects