Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.

270.0 mL of pure water

pHinitial, pHfinal

270.0 mL of a buffer solution that is 0.200 M in HCHO2 and 0.315 M in KCHO2

pHinitial, pHfinal

270.0 mL of a buffer solution that is 0.325 M in CH3CH2NH2 and 0.305 M in CH3CH2NH3Cl

pHinitial, pHfinal

Answer 1

1: For pure water, pH = 7.0

Moles of NaOH added = 0.005 mol

Volume of the solution, V = 270 mL = 0.270 L

Hence concentration of NaOH, [NaOH] = [OH-] = 0.005 mol / 0.270 L = 0.01852 M

Hence pOH = - log[OH-] = - log(0.01852 M) = 1.73

Hence final pH = 14 - pOH = 14 - 1.73 = 12.27 (answer)

Q.2: For HCOOH, pKa = 3.77

Hence initial pH of the buffer = pKa + log[KCHO2] / [HCHO2]

=> initial pH = 3.77 + log(0.315 M / 0.200M) = 3.97 (answer)

initial moles of HCHO2 = MxV = 0.200 mol/L x 0.270 L = 0.054 mol

initial moles of KCHO2 = MxV = 0.315 mol/L x 0.270 L = 0.08505 mol

When we add 0.005 mol NaOH it reacts with 0.005 mol HCOOH to form 0.005 mol salt KHCO2.

-------------- HCOOH + OH-(aq) ------ > CHO2-(aq) + H2O

Init.mol: 0.054 mol, 0.005 mol ---------- 0.08505 mol

change: - 0.005 mol, - 0.005 mol--------- +0.005 mol

mol.aft.rxn:0.049 mol, 0 mol ------------- 0.09005 mol

pH = pKa + log [CHO2-] / [HCOOH]

=> pH = pKa + log moles of CHO2- / moles of HCOOH

=> pH(final) = 3.77 + log(0.09005 mol / 0.049 mol) = 4.03 (answer)

Q.3:

For CH3CH2NH2, pKb = 3.19

Hence initial pOH of the buffer = pKb + log[CH3CH2NH3+] / [CH3CH2NH2]

=> initial pOH = 3.19 + log(0.305 M / 0.325M) = 3.16

=> Initial pH = 14 - pOH = 14 - 3.16 = 10.84 (answer)

initial moles of CH3CH2NH2 = MxV = 0.325 mol/L x 0.270 L = 0.08775 mol

initial moles of CH3CH2NH3+ = MxV = 0.305 mol/L x 0.270 L = 0.08235 mol

When we add 0.005 mol NaOH it reacts with 0.005 mol CH3CH2NH3+ to form 0.005 mol salt CH3CH2NH2 and H2O.

pOH = pKb + log [CH3CH2NH3+] / [CH3CH2NH2]

=> pOH = pKb + log (moles of CH3CH2NH3+ / moles of CH3CH2NH2)

=> pOH(final) = 3.19 + log(0.07735 mol / 0.09275 mol) = 3.11

=> pH (final) = 14 - 3.11 = 10.89 (answer)