Question

To 160.0 mL of a buffer solution containing 0.250 M ammonia and 0.100 M ammonium chloride, 20.00 mL of 1.00M HClO4 are added. a) Calculate the pH of the initial buffer solution. b) Calculate the pH of the new solution after the addition of acid. c) What becomes the pH if another 20.00 mL of the same acid solution is added? Pkb= 4.75

Answer 1

a).NH3= 160.0 ml of 0.250M

number of moles of NH3 = 0.250M x 0.160L= 0.04 moles

NH4Cl = 160.0 ml of 0.100M

number of moles of NH4Cl= 0.100M x 0.160L = 0.016 moles

PKb of NH3=4.75

for basic buffer

POH= PKb + log[salt]/[bae]

POH= 4.75 + log[0.016/0.04]

POH= 4.35

POH+ PH= 14

PH= 14-POH = 14-4.35 = 9.65

PH= 9.65

B)

HClO4 = 20.00ml of 1.00M

number of moles of HClO4 = 1.00M x 0.020L= 0.02 moles

after addition of Acid

number of moles of NH3 = 0.04 - 0.02 = 0.02 moles

number of moles of salt = 0.016 + 0.02 = 0.036 moles

POH= 4.75 + log[0.036/0.02]

POH= 5.00

PH= 14- 5.00= 9.00

PH= 9.00

c)

volume of acid = 20.00+ 20.00 = 40.00 ml

number of moles of acid = 1.0M x 0.040L = 0.04 moles

number of moles of acid = number of moles of NH3.

so it is the equivalent point

at equivalent point

PH= 7- 1/2[PKb + logC]

C = number of moles /total volume

total volume= 160.0+ 40.0= 200 ml= 0.2L

C= 0.04/0.2= 0.2M

PH= 7- 1/2[4.75 + log(0.2)]

PH= 4.97