Question

A solution is prepared by dissolving 15.0 g of pure HC2H3O2 and 25.0 g of NaC2H3O2 in 775 mL of solution (the final volume). (a) What is the pH of the solution? (b) What would the pH of the solution be if 25.0 mL of 0.250 M NaOH were added? (c) What would the pH be if 25.0 mL of 0.40 M HCl were added to the original 775 mL of buffer solution?

Answer 1

a)

Number of moles = mass/molar mass

Number of moles of HC2H3O2 = 15.0g / 60.03g/mol = 0.2499mol

Number of moles of NaC2H3O2 = 25.0g / 82.01g/mol = 0.3048mol

[HC2H3O2] = (0.2499mol/775ml)×1000ml = 0.3225M

[NaC2H3O2] = (0.3048mol/775ml)×1000ml = 0.3933M

pKa of CH3COOH = 4.75

Henderson - Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

pH = 4.75 + log(0.3933M/0.3225M)

pH = 4.75 + 0.09

pH = 4.84

b)

Number of moles of NaOH added = (0.250mol/1000ml)×25ml = 0.00625mol

NaOH react with weak acid HC2H3O2

HC2H3O2 + OH^- -------> C2H3O2^- + H2O

0.00625moles of NaOH react with 0.00625moles of HC2H3O2 to produce 0.00625 moles of C2H3O2^-

After addition of NaOH

Number of moles of HC2H3O2 = 0.2499mol - 0.00625mol = 0.2437

Number of moles of C2H3O2^- = 0.3048mol + 0.00625mol = 0.3111

Total volme = 775ml + 25ml = 800ml

[HC2H3O2] = (0.2437mol/800ml)×1000ml = 0.3046M

[C2H3O2^-] = (0.3111mol/800ml ) ×1000ml = 0.3889M

Applying H-H equation

pH = 4.75 + log(0.3889M/0.3046M)

pH = 4.75 + 0.11

pH = 4.86

c)

Number of moles of HCl added = (0.25mol/1000ml) ×40ml = 0.01mol

HCl react with conjucate base C2H3O2^-

C2H3O2^- + H⁺ -------> HC2H3O2

0.01moles of HCl react with 0.01moles of C2H3O2^- to produce 0.01moles of HC2H3O2

After addition of HCl

Number of moles HC2H3O2 = 0.2499mol + 0.01mol = 0.2599mol

Number of moles of C2H3O2^- = 0.3048mol - 0.01mol = 0.2948mol

Total volume =775ml + 40ml = 815ml

[HC2H3O2]= (0.2599mol/815ml)×1000ml = 0.3189M

[C2H3O2^-] = (0.2948mol/815ml )×1000ml = 0.3617M

Applying H-H equation

pH = 4.75 + log(0.3617M/0.3189M)

pH = 4.75 + 0.05

pH = 4.80