Question

A chemistry graduate student is given 300.mL of a 1.10M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with Ka=6.3x10^−5.

What mass of KC6H5CO2 should the student dissolve in the HC6H5CO2 solution to turn it into a buffer with pH =4.40?

You may assume that the volume of the solution doesn't change when the KC6H5CO2 is dissolved in it.

Be sure your answer has a unit symbol, and round it to 2 significant digits

Answer 1

We know that, pH of buffer solution is calculated by using Henderson's equation.

pH = pKa + log [ Salt ] / [ Acid ]

We have , pH = 4.40 , pKa = - log Ka = - log ( 6.3 10 ^-05 )

4.40 = - log ( 6.3 10 ^-05 ) + log [ C₆H₅COOK ] / [ C₆H₅COOH ]

4.40 = 4.20 + log [ C₆H₅COOK ] / [ C₆H₅COOH ]

log [ C₆H₅COOK ] / [ C₆H₅COOH ] = 4.40 - 4.20

log [ C₆H₅COOK ] / [ C₆H₅COOH ] = 0.20

[ C₆H₅COOK ] / [ C₆H₅COOH ] = 10 ^0.20

[ C₆H₅COOK ] / [ C₆H₅COOH ] = 1.58

[ C₆H₅COOK ] = 1.58 [ C₆H₅COOH ]

We can write , No. of moles of C₆H₅COOK = 1.58 No. of moles of C₆H₅COOH

Now, calculate no. of moles of C₆H₅COOH.

We know that, Molarity = No. of moles of solute / volume of solution in L

No. of moles of C₆H₅COOH = [ C₆H₅COOH ] volume of solution in L

No. of moles of C₆H₅COOH = 1.10 mol / L 0.300 L = 0.330 mol

No. of moles of C₆H₅COOK = 1.58 0.330 mol = 0.5214 mol

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

Mass = no. of moles molar mass

Molar mass of C₆H₅COOK = ( 7 12.01 ) + ( 5 1.0079 ) + ( 2 16.00 ) + 39.10 = 160.21 g / mol

Mass of C₆H₅COOK = 0.5214 mol 160.21 g /mol = 83.5 g

ANSWER : Student should add 84 g of of C₆H₅COOK to the 300 ml of 1.1 M C₆H₅COOH acid solution to turn it into buffer solution of pH 4.40.