Question

Titrating triprotic acid with strong base.Prepare a spreadsheet to graph the titration of 50.0 mL of 0.020 0 M histidine^.2HCl with 0.100 M NaOH. Treat histidine^.2HCl with the triprotic acid equation in Table 11-5

Short answer

The spreadsheet developed is shown below

The graph obtained from the above table is as follows

Step-by-step solution

Information given

Titration of 50.0 mL of 0.020 M histidine^.2HCl with 0.100 M NaOH.

Histidine.2HCl should be considered as a triprotic acid here and NaOH is a strong base. So, the reaction here will be in between triprotic acid with strong base.

Equation need to be used to develop spreadsheet

From table 11.5 the equation for the fraction of titration of triprotic acid with strong base is as follows

$\varphi =\frac{C_b{V}_b}{C_a{V}_a}=\frac{{\alpha }_{H_2{A}^{-}}+2{\alpha }_{H{A}^{2-}}+3{\alpha }_{A^{3-}}-\frac{\left[H^{+}\right]-\left[O{H}^{-}\right]}{C_a}}{1+\frac{\left[H^{+}\right]-\left[O{H}^{-}\right]}{C_b}}$

Other formulas need to be used

$\begin{array}{l}\left[H^{+}\right]={10}^{-pH}\\ \left[O{H}^{-}\right]=\frac{K_w}{\left[H^{+}\right]}\\ pK=-\log K\\ p{K}_1=1.6\\ p{K}_2=5.97\\ p{K}_3=9.28\end{array}$

$\begin{array}{c}\left[{\alpha }_{H_2{A}^{-}}\right]=\frac{{\left[H^{+}\right]}^2{K}_1}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3}\\ \left[{\alpha }_{H{A}^{2-}}\right]=\frac{\left[H^{+}\right]K_1{K}_2}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3}\\ \left[{\alpha }_{A^{3-}}\right]=\frac{K_1{K}_2{K}_3}{{\left[H^{+}\right]}^3+{\left[H^{+}\right]}^2{K}_1+\left[H^{+}\right]K_1{K}_2+K_1{K}_2{K}_3}\end{array}$

Spreadsheet

Spreadsheet developed

Graph Plot

The graph obtained from the above spreadsheet is depicted below