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

Answer:

$3.03\times 10^{25}$ particles of $A^-$ are in the container after the neutralization.

Explanation:

Chemical equation for the neutralization reaction:

$HA+NaOH\rightarrow H_2O+NaA$

Number of particles of HA dissolved in water ,x= $3.9\times 10^{25}$

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

$y=8.7\times 10^{24}$

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

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

$x-y=3.9\times 10^{25}-8.7\times 10^{24}=3.03\times 10^{25}$