Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH

(For all answers express answers using two decimal places seperated by a comma)

A) For 250.0 mL of pure water, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH

B) For a 250.0 mL of a buffer solution that is 0.235 M in HCHO2 and 0.275 M in KCHO2, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH ( Ka= 1.8x10^-4)

C) For 250.0 mL of a buffer solution that is 0.2816 M in CH3CH2NH2 and 0.2571 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH (Ka=5.6x10^-4)

Answer 1

A) initial pH of pure water = 7.0

final pH

moles of NaOH=0.01 mol

volume of solution = 250 mL =0.25L

Molarity = 0.01mol /0.25 L=0.04 M

Thus [OH-] = 0.04 M and pOH = -log0.04=1.3979

and pH = 14-pOh = 14-1.3979= 12.60 1

B)

HCHO2 +OH- _----------------> CHO2- + H2O

0.235 x250 0 0.275 x250 0 initial mmoles

=58.75 = 68.75

----------- 0.01mol = 10 - - ----- change

48.75 0 78.75 after addition

pKa of acid = -log 1.8x10^-4 = 3.75

Now

pH of buffer is givenby Hendersen equation as

pH = pKa + log [conjugate base]/[acid]

ph before addition = 3.75 + log 68.75/58.75=3.818

pH after addition = 3.75 + log 78.75 /48.75 =3.958

C)

C2H5NH3Cl (acid) + OH- --------------------->C2H5NH2 (conjugate base)

0.2571 x250 0 0.2816x250 initial mmoles

= 64.275 =70.4

----------- 0.01mol = 10 ---       change

  54.275 0 80.4   after addition

pKa of acid = -log 5.6x10^-4 = 3.252

Now

pH of buffer is givenby Hendersen equation as

pH = pKa + log [conjugate base]/[acid]

ph before addition = 3.25 + log 70.4//64.275=3.29

pH after addition = 3.25 + log 80.4/54.275 =3.42