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

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.

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

$0.3364 * 160 =mass \ of \ KC_6H_5CO_2$

$mass \ of \ KC_6H_5CO_2 = 54 g$

Answer from: Quest

A. volume stays constant. just took took the quiz

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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A chemistry graduate student is given of a benzoic acid solution. Benzoic acid is a weak acid with ....