Question

23.For a 0.1Maqueous solution of sodium acetate,$N{a}^{+}{\mathrm{H}}_3{\mathrm{CO}}_2^{-}$, one mass alance is simply$\left[{\mathrm{Na}}^{+}\right]=0.1\mathrm{M}$. Write a mass balance involving acetate.

Short answer

The mass balance involving acetate is given by $\left[{\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}\right]+\left[{\mathrm{CH}}_3{\mathrm{CO}}_2^{-}\right]=0.1\mathrm{M}$.

Step-by-step solution

Definition of mass balance

The mass balance states that the quantity of all species in a solution containing a particular atom (or group of atoms) must equal the amount of that atom (or group) delivered to the solution.

Balancing the mass for given solution

Read and understand the following to come up with a mass balance

involving acetate:

(1)If the acetate does not react any further,$\left[{\mathrm{CH}}_3{\mathrm{CO}}_2^{-}\right]=0.1\mathrm{M}$

(2) However,1 mole of the ${\mathrm{CH}}_3{\mathrm{CO}}_2^{-}$reacts with water to form 1

mole of${\mathrm{CH}}_3{\mathrm{CO}}_2^{-}$.

So, the mass balance is $\left[{\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}\right]+\left[{\mathrm{CH}}_3{\mathrm{CO}}_2^{-}\right]=0.1\mathrm{M}$.

Therefore, the mass balance involving acetate is given by

$\left[{\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}\right]+\left[{\mathrm{CH}}_3{\mathrm{CO}}_2^{-}\right]=0.1\mathrm{M}$.