Question

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 solution to turn it into a buffer with pH ? You may assume that the volume of the solution doesn't change when the is dissolved in it. Be sure your answer has a unit symbol, and round it to significant digits.

Answer 1

Complete Question

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with
`pKa = 4.2`. What mass of KC6H5CO2 should the student dissolve in the HC6H5CO2 solution to turn it into a buffer with pH =4.63? 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:

The mass would be
`mass \ of \ KC_6H_5CO_2 = 54 g`

Step-by-step explanation:

From the question we are told that

concentration of benzoic acid is
`M_B = 1M`

The volume of benzoic acid is
`V_B = 125mL = 0.125L`

The value of the experimental parameter is
`pKa = 4.2`

The value of PH is
`PH =4.63`

Generally the number of moles is mathematically represented as

No of moles = concentration * volume

`N = M_B * V_B`

Substituting values

`N = 1 * 0.125 = 0.125\ moles`

Generally PH is mathematically represented as

`PH = pKa + log([KC_6H_5CO_2])/([HC_6H_5CO_2])`

Where
`[KC_6H_5CO_2]` is the No of moles of
`KC_6H_5CO_2` and
`[HC_6H_5CO_2]` is the number of moles of benzoic acid

`4.3 = 4.2 + log([KC_6H_5CO_2])/(0.125)`

`4.63 - 4.2 = log ([KC_6H_5CO_2])/(0.125)`

`0.43 =log ([KC_6H_5CO_2])/(0.125)`

`10^(0.43) =([KC_6H_5CO_2])/(0.125)`

`2.6915 *0.125 = [KC_6H_5CO_2]`

`[KC_6H_5CO_2]= 0.3364`

Generally the number of moles is mathematically represented as

`No \ of \ moles \ = (Mass)/(Molar\ Mass)`

For
`[KC_6H_5CO_2]`

`0.3364 = (mass \ of \ KC_6H_5CO_2)/(Molar \ Mass \ of \ KC_6H_5CO_2 )`

Molar mass of
`KC_6H_5CO_2` = 160g/mole

`0.3364 * 160 =mass \ of \ KC_6H_5CO_2`

`mass \ of \ KC_6H_5CO_2 = 54 g`