Question

1. Find the pH of a solution prepared from 0.392 g of potassium acetate (CH3CO2K) and 15.00 mL of 0.0470 M of acetic acid (CH3CO2H) diluted to 100.00 mL.

Answer 1

Solution contain weak acid acetic acid and its salt potassium acetate. This solution acts as a buffer solution . To find out pH of this solution , we need to calculate [ Acetic acid ] & [ potassium acetate ] in the diluted 100 ml solution.

We can calculate [ Acetic acid]  by using dilution formula.

We have formula, C _stock$\times$ V _stock = C _dilute  $\times$ V _dilute

$\therefore$ C _dilute = C _stock$\times$ V _stock / V _dilute

$\therefore$ C _dilute = 0.0470 M $\times$ 15.00 ml / 100.00 ml = 0.00705 M

Hence, [ Acetic acid ] = 0.00705 M

Now, calculate the [ Potassium acetate ]

Mass of potassium acetate = 0.392 g

Molar mass of CH₃CO₂K = ( 2 $\times$ 12.01 ) + ( 3 $\times$ 1.0079 ) + ( 2 $\times$ 16.00 ) + 39.10 = 98.14 g / mol

We have , No. of moles = Mass / molar mass

$\therefore$ No. of moles of CH₃CO₂K = 0.392 g / 98.14 g /mol = 0.00399 mol

Volume of solution = 100 ml = 0.100 L

[ CH₃CO₂K ] = No. of moles of CH₃CO₂K / Volume of solution in L

[ CH₃CO₂K ] = 0.003994 mol / 0.100 L = 0.0399 M

p H of buffer solution is calculated by using equation , pH = pKa + log [ Salt ] / [ Acid ]

$\therefore$ pH = - log Ka + log [ CH₃CO₂K ] / [ CH₃CO₂H ]

$\therefore$ pH = - log ( 1.75 $\times$ 10 ^-05 ) + log 0.0399 / 0.00705

pH = 4.757 + log 0.0399 / 0.00705

pH = 4.757 + 0.753

pH = 5.51

ANSWER : pH of solution = 5.51