Question

Write two mass balances for a 1.0Lsolution containing 0.100molof sodium acetate.

Short answer

The mass balances are$\left[{\mathrm{Na}}^{+}\right]=0.100\mathrm{M}$ and $\left[{\mathrm{CH}}_3{\mathrm{CO}}_2^{-}\right]+\left[{\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}\right]=0.100\mathrm{M}$.

Step-by-step solution

Concept used.

Mass balance:

All species in a solution containing a certain atom (or group of atoms) must be equal to the amount of that atom (or group) given to the solution.

Step 2: Calculate the two mass balances.

By definition, mass balance of the solution is given by

For sodium ion:

$\left[{\mathrm{Na}}^{+}\right]=0.100M$

For acetate ion:

$$\begin{gathered} \left[{\mathrm{CH}}_3{\mathrm{CO}}_2^{-}\right]+\left[{\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}\right]=0.100\mathrm{M} \\ \left(\mathrm{dissociated}\mathrm{product}\right)\left(\mathrm{undissociated}\mathrm{product}\right)\left(\mathrm{amount}\mathrm{of}\mathrm{acetic}\mathrm{acid}\mathrm{put}\mathrm{into}\mathrm{the}\mathrm{solution}\right) \end{gathered}$$

Therefore, mass balance equation for 1.0L solution of $0.100\mathrm{molNa}+{\mathrm{CH}}_3{\mathrm{CO}}_2^{-}$was given.