Question

Calculate the pH of the solution that results from each of the following mixtures.

a) 140.0 mL of 0.24 M HF with 220.0 mL of 0.30 M NaF

b) 165.0 mL of 0.12 M C2H5NH2 with 285.0 mL of 0.22 M C2H5NH3Cl

Express your answer using two decimal places.

Answer 1

HF is weak acid and NaF is the   salt of weak base. F- is conjugate base.

Henderson–Hasselbalch equation can be written as

pH= pKa+ log [F-]/[HF]

where [F-]= conjugate base of HF( NaF suppliments this)

Ka= dissociation constant for HF, pKa= 3.17

140.0 mL of 0.24 M HF with 220.0 mL of 0.30 M NaF

Moles of HF in 140ml of 0.24M= Molarity* Volume (L)= 0.24*140/1000=0.0336

Moles of NaF in 220ml of 0.30M= 0.3*0.22= 0.066

Volume of solution after mixing= 140+220=360ml=0.36L, Concentrations after mixing : HF= 0.0336/0.36 and NaF= 0.066/0.36

pH= 3.17+ log (0.066/0.0336)= 3.46

2. C2H5NH2 is base and C2H5NH3Cl is salt of of the base and C2H5NH3+ is conjugate acid.

Henderson–Hasselbalch equation can be written as

POH = pKb + log [C2H5NH3+]/[C2H5NH2] , pKb for C2H5NH2= 3.25

moles of C2H5NH2 in 165 ml of 0.12M= Molarity* Volume (L)=0.12*0.165= 0.0198, moles of C2H5NH3+ =0.22*0.285= 0.0627, Volume of the solution after mixing = 285+165= 450 ml=0.45L

Concentrations: C2H5NH2= 0.0198/0.45 and C2H5NH3+ =0.0627/0.45

pOH= 3.25+ log ( 0.0627/0.0198)= 3.75, pH= 14-pOH= 14-3.75= 10.25