Question

question 5

For each of the following solutions, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Part B

250.0 mL of a buffer solution that is 0.200 M in HCHO2 and 0.305 M in KCHO2

Express your answers using two decimal places separated by a comma.

Part C

250.0 mL of a buffer solution that is 0.265 M in CH3CH2NH2 and 0.245 M in CH3CH2NH3Cl

Express your answers using two decimal places separated by a comma.

Answer 1

Part-B

HCHO2= 250mL of 0.200M

number of moles of HCHO2 = 0.200M x0.250L= 0.05 moles

KCHO2= 250ml of 0.305M

Number of moles of KCHO2 = 0.305M x 0.250L = 0.07625 noles

PKa of HCHO2= 3.8

PH= PKa +log[salt]/[acid]

PH= 3.8 + log[0.07625/0.05]

PH= 3.98
Initial PH= 3.98

Number of moles of NaOH= 0.010 mol

after addition of NaOH

number of moles of HCHO2= 0.05 - 0.010= 0.04 moles

number of moles of KCHO2= 0.07625 + 0.010 = 0.08625

PH= 3.8 + log[0.08625/0.04]

PH= 4.13

Part-C

CH3CH2NH2 = 250ml of 0.265M

number of moles of CH3CH2NH2= 0.265 x 0.250 = 0.06625 moles

CH3CH2NH3Cl = 250 ml of 0.245M

number of moles CH3CH2NH3Cl= 0.245M x 0.25L= 0.06125 moles

Kb= 4.3x10^-4

-log[Kb] = -log[4.3x10^-4]

PKb= 3.37

POH= PKb + log[salt/base]

POH= 3.37 + log[0.06125/0.06625]

POH= 3.36

POH + POH= 14

PH= 14-POH= 14 -3.36= 10.64

PH= 10.64

number of moles of NaOH= 0.010 mole

after additon of NaOH

number of moles CH3CH2NH2 = 0.06625 + 0.010= 0.07625 moles

number of moles of CH3CH2NH3Cl = 0.06125 - 0.010= 0.05125 mole

POH = 3.37 + log[0.05125/0.07625]

POH= 3.2

PH= 14-POH= 14-3.2= 10.8

PH= 10.8