91Ó°ÊÓ

• The Henderson Hasselbalch equation is a simplified formula that shows the connection between a solution's pH or pOH, pKa or pKb, and the ratio of concentrations of dissociated chemical species.
• One of the most significant economically, sulfuric acid (H₂SO₄), also known as oil of vitriol or hydrogen sulphate, is a viscous, colourless, oily, caustic liquid.

This will calculate the mass of ${\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{FM}=142.04\mathrm{g}/\mathrm{mol}\right)$sulfuric acid should be supplemented with this. (FM=98.08g/mol) in order to give 1Lwith a buffer pH=2.80 with a total sulphur content of 0.2M

$\left(={\mathrm{SO}}_4^{2-}+{\mathrm{HSO}}_4^{-}+{\mathrm{H}}_2{\mathrm{SO}}_4\right)$

It'll start with the Henderson-Hasselbach equation yields the following:

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}++\log \left(\left[[\mathrm{A}-1\right]/\left[\mathrm{HA}\right]\right)$

$2.0=1.987+\log \frac{\left[{\mathrm{SO}}_4^{2-}\right]}{\left[{\mathrm{HSO}}_4^{-}\right]}$

$\left[{\mathrm{HSO}}_4^{-}\right]=0.1538\left[{\mathrm{SO}}_4^{2-}\right]$

Consider the response.

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_4+{\mathrm{SO}}_4^{2-}2{\mathrm{HSO}}_4^{-} \\ \left[{\mathrm{SO}}_4^{2-}\right]=\mathrm{y}-\mathrm{x} \end{gathered}$$

$\left[{\mathrm{HSO}}_4^{-}\right]=2\mathrm{x}$

it should be noted that ${\mathrm{H}}_2{\mathrm{SO}}_4$is x at the start and${\mathrm{SO}}_4^{2-}$is y at first

It may write the following using the result from above step and combining it with $$\begin{gathered} \left[{\mathrm{HSO}}_4^{-}\right]=0.1538\left[{\mathrm{SO}}_4^{2-}\right] \\ 2\mathrm{x}=01583\left(\mathrm{y}-\mathrm{x}\right) \\ \mathrm{x}+\mathrm{y}=0.2\mathrm{mol}\mathrm{from}\mathrm{the}\mathrm{task} \\ \mathrm{x}-0.2=\mathrm{y} \end{gathered}$$

$$\begin{gathered} 2\left(0.2\mathrm{y}\right)=0.1538\left(\mathrm{y}-\left(0.2-\mathrm{y}\right)\right) \\ \mathrm{y}=0.1867\mathrm{mol} \\ \mathrm{x}=0.0133\mathrm{mol} \end{gathered}$$

The mass of the following can then be calculated:

$$\begin{gathered} m\left({\mathrm{Na}}_2{\mathrm{SO}}_4\right)=\mathrm{n}×\mathrm{M} \\ m\left({\mathrm{Na}}_2{\mathrm{SO}}_4\right)=0.1867\mathrm{mol}×142.04\mathrm{g}/\mathrm{mol} \\ m\left({\mathrm{Na}}_2{\mathrm{SO}}_4\right)=26.52\mathrm{g} \\ m\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=\mathrm{n}×\mathrm{M} \\ m\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=0.0133\mathrm{mol}×98.08\mathrm{g}/\mathrm{mol} \\ m\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=0.131\mathrm{g} \end{gathered}$$

The mass of${\mathrm{Na}}_2{\mathrm{SO}}_4$to be added is 1.31 grams.