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Assume a container starts with 3.9 x1025 HA particles dissolved in water and none of its conjugate. Sodium hydroxide is then added to neutralize the acid until the number of HA particles = 8.7x1024. Write out the chemical equation for the neutralization reaction and calculate the number of A‐ particles that are in the container after the neutralization.

Respuesta :

Answer:

[tex]3.03\times 10^{25}[/tex] particles of [tex]A^-[/tex] are in the container after the neutralization.

Explanation:

Chemical equation for the neutralization reaction:

[tex]HA+NaOH\rightarrow H_2O+NaA[/tex]

Number of particles of HA dissolved in water ,x= [tex]3.9\times 10^{25}[/tex]

Single molecule of HA had single [tex]H^+[/tex] and 1 [tex]A^-[/tex].

So , Number of particles of [tex]A^-[/tex] dissolved in water = x

Number of particles HA which get neutralized by adding NaOH : y

[tex]y=8.7\times 10^{24}[/tex]

Number of [tex]A^-[/tex] particles get neutralized by adding NaOH = y

Number of [tex]A^-[/tex] particles that are in the container after the neutralization:

[tex]x-y=3.9\times 10^{25}-8.7\times 10^{24}=3.03\times 10^{25}[/tex]

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