A chemistry graduate student is given of a benzoic acid solution. Benzoic acid is a weak acid with . What mass of should the student dissolve in the s

53.9g of KC₆H₅CO₂ is the mass the student should dissolve.

Explanation:

Full question is:

A chemistry graduate student is given 125.mL of a 1.00M 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 =4.63?

It is possible to find pH of a buffer by using H-H equation:

pH = pka + log₁₀ [A⁻] / [HA]

Where pKa is -log Ka (4.2), [A⁻] is concentration of conjugate base, KC₆H₅CO₂ and [HA] concentration of the weak acid, HC₆H₅CO₂

-This concentration could be replaced per moles of each compound-

Moles of 125mL = 0.125L of a 1.00M of benzoic acid are:

0.125L × (1.00mol / L) = 0.125 moles benzoic acid.

Replacing in H-H equation, for a pH of 4.63:

4.63 = 4.2 + log₁₀ [A⁻] / [0.125mol]

0.43 = log₁₀ [A⁻] / [0.125mol]

2.6915 = [A⁻] / [0.125mol]

0.336 moles = [A⁻]

That means the student needs 0.336 moles of KC₆H₅CO₂ to obtain the buffer with pH = 4.63.

As molar mass of KC₆H₅CO₂ is 160.21g/mol, mass of 0.336 moles are:

0.336 moles × (160.21g / mol) =