Chemistry, 16.04.2020 04:13 emarquez05

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.

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Answer from: melanis52

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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.

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

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

$PH = pKa + log\frac{[KC_6H_5CO_2]}{[HC_6H_5CO_2]}$

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

$4.3 = 4.2 + log\frac{[KC_6H_5CO_2]}{0.125}$

$4.63 - 4.2 = log \frac{[KC_6H_5CO_2]}{0.125}$

$0.43 =log \frac{[KC_6H_5CO_2]}{0.125}$

$10^{0.43} =\frac{[KC_6H_5CO_2]}{0.125}$

2.6915 *0.125 = [KC_6H_5CO_2]

[KC_6H_5CO_2]= 0.3364

Generally the number of moles is mathematically represented as

$No \ of \ moles \ = \frac{Mass}{Molar\ Mass}$

For [KC_6H_5CO_2]

$0.3364 = \frac{mass \ of \ KC_6H_5CO_2}{Molar \ Mass \ of \ KC_6H_5CO_2 }$

Molar mass of KC_6H_5CO_2 = 160g/mole

$0.3364 * 160 =mass \ of \ KC_6H_5CO_2$

$mass \ of \ KC_6H_5CO_2 = 54 g$

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