Buffers thanks Your assigned pH is 46 You are to assume that

Buffers

thanks

Your assigned pH is: 4.6 You are to assume that the buffer is made up by mixing volumes of 0.100 M acetic acid and 0.100 M sodium acetate solutions Calculate the volume of 0.100 M acetic acid required to prepare 60.0 mL of a buffer of pH 4.6 35.0 mL 1 pts You are correct. Previous Tries Your receipt no. is 157-4962 Calculate the volume of 0.100 M sodium acetate required to prepare 60.0 mL of a buffer of pH 4.6. 25.0 mL 1 pts You are correct. Previous Tries Your receipt no. is 157-3996 Calculate the concentration [HA] in this buffer. 0.0717 M 1 pts Submit Answer Incorrect. Tries 4/99 Previous Tries Calculate the concentration [A] in this buffer 0.0717 M 1 pts Submit Answer Incorrect. Tries 1/99 Previous Tries

Solution

1)

Answer

[HA] =0.0586M

[A-]= 0.0414M

Explanation

According to Henderson - Hasselbalch

pH = pKa + log([A-]/[HA])

Buffer pH = 4.6

pKa of Acetic acid = 4.75

Therefore,

4.6 = 4.75 + log([A-] /[HA])

log([A-] /[HA]) = - 0.15

[A-] /[HA] = 0.7079

[A-] = 0.7079[HA]

total buffer concentration is 0.100M

So,

0.7079[HA] + [HA] = 0.100M

1.7079[HA]= 0.100M

[HA] = 0.0586M

[A-] = 0.100M - 0.0586M = 0.0414M

2)

Answer

Volume of 0.1M NaOH for base equivalence point =11.72ml

pH at base equivalence point = 8.77

Explanation

i) NaOH react with HA

NaOH + HA - - - ->NaA + H2O

stoichiometrically, 1mole of NaOH react with 1mole of HA

No of mole of HA = (0.0586mol/1000ml)×20ml = 0.001172

0.001172moles of HA react with 0.001172moles of NaOH

Volume of 0.1M NaOH solution Containing 0.001172mol of NaOH =(1000ml/0.1000mol)×0.001172mol =11.72ml

ii) At base equivalence point all the HA is converted into A-

A- is partly hydrolysed by water

A- + H2O <- - - - - > HA + OH-

Kb= [HA] [OH-] /[A-]

Total volume at base equivalence point = 20ml +11.72ml =31.72ml

Diltion factor of A- concentration = 31.72/20= 1.586

[A-] = 0.100M/1.586 = 0.0631M

Kb = Kw/Ka = 1.00×10^-14/1.78×10^-5 = 5.62×10^-10M

at equillibrium,

x²/0.0631-x = 5.62×10^-10

  we can assume 0.0631 - x~0.0631

x²/0.0631 = 5.62×10^-10

x² = 3.54×10^-11

x = 5.95×10^-6

[OH-] = 5.95×10^-6M

pOH = 5.23

pH =14 - pOH

= 14 - 5.23

= 8.77

3)

Anaswer

Volume 0.1M HCl required for acid equivalence point = 8.28ml

pH at acid equivalence point = 2.95M

Explanation

i) HCl react with conjucate base A- and this reaction is 1:1 molar reaction

therefore,

V1 × M1 = V2 × M2

V2 = V1×M1/M2

= 20ml × 0.0414M/0.1M

= 8.28ml

ii) At acid equivalence point all the A- is converted into HA

total volume = 20 + 8.28 = 28.28ml

Dilution factor of concentration of HA = 28.28/20 =1.414

[HA] = 0.100M/1.414 = 0.0707M

Dissociation of HA is

HA <- - - - - >H+ + A-

Ka = [H+] [A-] /[HA]

at equillibrium

[HA] = 0.0707-x

[A-] = x

[H+] = x

Therefore,

x²/0.0707 - x = 1.78×10^-5

we can assume 0.0707-x ~ 0.0707

x²/0.0707 = 1.78×10^-5

   x² = 1.26×10^-6

     x = 1.12×10^-3

So,

[H+] = 1.12×10^-3M

pH = - log[H+]

= - log(1.12×10^-3)

= 2.95