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For the titration of 20.00 mL of 0.150 M HF with0.250 M NaOH, calculate: a) the...

For the titration of 20.00 mL of 0.150 M HF with0.250 M NaOH, calculate:

a) the initial pH;Ka= 6.6 x 10^-4

b) the pH when the neutralization is 25% complete

c) the pH when the neutralization is 50% complete

d) the pH when the neutralization is 100% complete

Solutions

Expert Solution

millimoles of HF = 20 x 0.150 = 3

concentration of NaOH = 0.250

Ka = 6.6 x 10^-4

pKa = 3.18

a) initial pH :

[H+] = sqrt (Ka x C)

= sqrt (6.6 x 10^-4 x 0.150)

= 9.95 x 10^-3 M

pH = -log [H+] = -log (9.95 x 10^-3)

= 2.00

pH = 2.00

b) the pH when the neutralization is 25% complete :

at this millimoles of NaOH = 3 / 4 = 0.75

HF + NaOH ---------------> NaF + H2O

3 0.75 0 0

2.25 0 0.75

pH = pKa + log [salt / acid]

= 3.18 + log [0.75 / 2.25]

= 2.70

pH = 2.70

c) the pH when the neutralization is 50% complete :

here pH = pKa

pH = 3.18

d) the pH when the neutralization is 100% complete :

here salt only remains :

volume at eqivalence point = 3 / 0.250 = 12.

salt concentration = millimoles / total volume

= 3 / (20 + 12)

= 0.09375 M

pH = 7 + 1/2 (pKa + log C)

= 7 + 1/2 (3.18 + log 0.09375)

= 8.08

pH = 8.08

queen_honey_blossom answered 1 year ago

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