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8090 [49]
3 years ago
14

: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne

t ionic equation for the reaction that occurs when NaOH is added to a buffer containing H2C6H5O7 - and HC6H5O7 2- ?
Chemistry
2 answers:
olchik [2.2K]3 years ago
8 0

Answer:

The Net reaction is    

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

Explanation:

From the Question we are told that the buffers are

                H_2C_6H_5O_7^ - and  HC_6H_5O_7 ^{ 2-}

When NaOH is added the Net ionic reaction would be

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

             

Mariana [72]3 years ago
7 0

Answer:

Net ionic equation are as follows

(1) H_3C_6H_5O_7 + 0H^- \rightarrow H_2C_6H_5O_7^- +H_2O \\\\ (2)  H_2C_6H_5O_7^ - + OH^- \rightarrow H_C_6H_5O_7^{ 2-} + H_2O \\\\  (3)  HC_6H_5O_7 ^{2-} + OH^- \rightarrow C_6H_5O_7^{3-} +H_2O

Explanation:

Citric acid, H3C6H5O7, is a triprotic acid

he net ionic equation for the reaction that occurs when NaOH is added to a buffer containing H2C6H5O7 - and HC6H5O7 2-

Net ionic equation are as follows

(1) H_3C_6H_5O_7 + 0H^- \rightarrow H_2C_6H_5O_7^- +H_2O \\\\ (2)  H_2C_6H_5O_7^ - + OH^- \rightarrow H_C_6H_5O_7^{ 2-} + H_2O \\\\  (3)  HC_6H_5O_7 ^{2-} + OH^- \rightarrow C_6H_5O_7^{3-} +H_2O

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vladimir2022 [97]

Answer:

C_{7} H_{5}N_{3}O_{6}

Explanation:

First reaction gives you the number of moles or the mass from Carbon and hydrogen

for carbon:

0,213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} =0.005molC

0.213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} .\frac{12gC}{1molC} = 0.058gC\\

Analogously for hydrogen:

0.0310gH_{2}O have 0.0034gH or 0.0034mol of H

In the second reaction you can obtain the amount of nitrogen as a percentage and find the mass of N in the first sample.

0.023gNH_{3} .\frac{1molNH_{3}}{17gNH_{3}} .\frac{1molN}{1molNH_{3}} .\frac{14gN}{1molN} \frac{100}{0.103gsample} =18.4%N

now

\frac{18.4gN}{100gsample} .0.157gsample=0.0289gN in the first reaction

this is equivalet to 0.002mol of N

with this information you can find the mass of oxygen by matter conservation.

gO=total mass-(gN+gC+gH)=0.157-(0.0289+0.058+0.0034)=0.0666gO

this is equivalent to 0.004molO

finally you divide all moles obtained between the smaller number of mole (this is mol of H)

C\frac{0.0048}{0.0034} H\frac{0.0034}{0.0034} N\frac{0.002}{0.0034} O\frac{0.004}{0.0034} =C_{1.4} HN_{0.6} O_{1.2}

and you can multiply by  5   to obtain: C_{7} H_{5}N_{3}O_{6}

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Answer:

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Explanation:

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q=mc\bigtriangleup T\\\\c=\frac{q}{m\bigtriangleup T}\\\\\bigtriangleup T=(57-32)\textdegree C=25\textdegree C\\\\\therefore c=\frac{6.75\times 10^4J}{1.3\times 1000\ g\times 25\textdegree  C}\\\\=2.0769 \ \frac{J}{g\ \textdegree C}

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