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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.

User Hanan
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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

User Jcbvm
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