Question

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations

a. 100.0 ml of 0.70M HC7H5O2 (Ka= 6.4x10^-5) titrated by 0.10 M NaOH

pH at the halfway point = ______?

pH at the equivalence point = _____?

b. 100.0ml of 0.70M C2H5NH2 (Kb= 5.6x10^-4) titrated by 0.60M HN03

pH at the halfway point = ______?

pH at the equivalence point = _____?

c. 100.0 ml of 0.70M HCL titrated by 0.15m NaOH

pH at the halfway point = ______?

pH at the equivalence point = _____?

Answer 1

a) 100.0ml of 0.70M HC7H5O2

HC7H5O2 = 100.0ml of 0.70M

number ofmoles of HC7H5O2 = 0.70Mx0.100L = 0.07 moles

Ka= 6.4x10^-5

-log(ka) = -log(6.4x10^-5)

Pka= 4.19

At half way point

PH= Pka

PH= 4.19

NaOH= 0.10M

at equivalent point

Volume of NaOH = 0.70x100/0.10= 700 mL

number of moles of NaOH = 0.10Mx0.700L = 0.07 moles

At equivalent point number ofmoles are equal

Total volume= 100+700=800mL= 0.800L

at equivalent point

PH= 7+1/2[PKa+logC]

C= number of moles/volume = 0.07/0.800=0.0875 M

PH = 7+1/2[4.19+log(0.0875)]

PH= 8.566= 8.57

PH= 8.57

b) C2H5NH2 = 100.0mL of 0.70M

number of moles of C2H5NH2 = 0.70Mx0.100L = 0.07 moles

Kb= 5.6x10^-4

-log(Kb) = -log(5.6x10^-4)

PKb = 3.25

at halfway point

POH= PKb

POH= 3.25

PH+POH=14

PH=14-POH

PH= 14-3.25

PH=10.75

HNO3=0.60M

at equivalent point

volume ofHNO3 = 0.70x100.0/0.60 =116.67 mL

number of moles ofHNO3 = 0.60Mx0.11667L= 0.070 moles

Total volume = 100.0+116.67= 216.67 mL= 0.21667L

at equivalent point

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

C= 0.07/0.21667=0.323M

POH= 7-1/2[3.25+log(0.323)]

POH= 5.62

PH= 14-5.62

PH= 8.38

c)