Question

500 mg of sodium acetate is dissolved in 500.0 mL of de-ionized water. Calculate the pH of this solution. MM(sodium acetate) = 82.0324 g/mol, K_a(acetic acid) = 1.75 x 10^-5

Answer 1

For weak acids (no 100% ionization) the relation for pH can be given as:

pH = 7 + 1/2[pka - logc] ------------------(1)

where in this case pka of acetic acid = -logKa = -log1.75 x 10^-5

pKa = - (-4.75)

pKa = 4.75

Now calculate the mole of sodium acetate = 0.5g/82.0324 = 0.00609 mole

Now, concentration of sodium acetate "c" in 500 mL water = 0.00609 mole x 1000/500 M

= 1.218 x 10^-2M

put all these values in above equation (1):

pH = 7 + 1/2[4.75 - log1.218 x 10^-2]

pH = 7 + 1/2[4.75 -1.89]

pH = 7 + 1.43

pH = 8.43