Question

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.

Answer 1

`3.03* 10^(25)` particles of
`A^-` are in the container after the neutralization.

Step-by-step explanation:

Chemical equation for the neutralization reaction:

`HA+NaOH\rightarrow H_2O+NaA`

Number of particles of HA dissolved in water ,x=
`3.9* 10^(25)`

Single molecule of HA had single
`H^+` and 1
`A^-`.

So , Number of particles of
`A^-` dissolved in water = x

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

`y=8.7* 10^(24)`

Number of
`A^-` particles get neutralized by adding NaOH = y

Number of
`A^-` particles that are in the container after the neutralization:

`x-y=3.9* 10^(25)-8.7* 10^(24)=3.03* 10^(25)`