Question

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

1. For 240.0 mL of pure water, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

2. For 240.0 mL of a buffer solution that is 0.195M in HCOOH and 0.280 M in HCOOK, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

3. For 240.0 mL of a buffer solution that is 0.255 M in C2H5NH2 and 0.225 M in C2H5NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Comment

Answer 1

1) pH of pure water is 7

[OH-]= 0.010mol/0.240L= 0.042M

pOH= -log [OH-]= -log 0.042M= 1.38 ---> pH + pOH= 14 ---> pH= 14-1.38= 12.62

2) pH= pKa + log[HCOO-]/[HCOOH]

pKa=3.74

pH= 3.74 + log 0.28/0.195= 3.9 (initial)

when we add NaOH the OH - will react with the HCOOH, so the amount of HCOOH will decrease by 0.01mol and the amount of HCOO- will increase by 0.01mol.

mol of HCOOH= 0.195M x 0.24L= 0.0468 mol

mol of HCOO- = 0.28M x 0.24L= 0.0672 mol

Final moles:

mol of HCOOH= 0.0468 mol - 0.01mol= 0.0368 mol -----> [HCOOH]= 0.0368mol/0.24L= 0.153M

mol of HCOO- =0.0672 + 0.01mol= 0.0772 mol ------> [HCOO-]= 0.0772mol/0.24L= 0.322M

pH= 3.74 + log 0.322/0.153 = 4.06 (final)

3) pKb= 3.2 (this is the data i have, maybe you have another data and the result can vary a little)

pOH= pKb + log [C2H5NH3+]/[C2H5NH2]

pOH= 3.2 + log 0.225/0.255 =3.15----> pH + pOH=14 ----> pH= 14-3.15= 10.85 (initial)

the 0.01 mol of OH- will react with C2H5NH3+, so final moles will be:

moles of C2H5NH3+= (0.225M x 0.24L) -0.01mol= 0.044mol ---> [C2H5NH3+]= 0.044mol/0.24L= 0.183M

moles of C2H5NH2=(0.255M x 0.24L) + 0.01mol= 0.0712 mol ----> [C2H5NH2]= 0.0712mol/0.24L= 0.297M

pOH= 3.2 + log 0.183/0.297 = 2.99 ---> pH + pOH= 14 ----> pH= 14-2.99= 11.01