91影视

50.0 mL of 0.020 0 M H₂A needs to be titrated with 0.100 M NaOH

. (pK₁=4.00, pK₂=8.00)

Fromtable 11.5 the equation for the fraction of titration of diprotic acid (H₂A) with strong base (B) is as follows

$蠒=\frac{C_b{V}_b}{C_a{V}_a}=\frac{{伪}_{H{A}^{-}}+2{伪}_{A^{2-}}-{\displaystyle \frac{\left[H^{+}\right]-\left[O{H}^{-}\right]}{C_a}}}{1+\frac{\left[H^{+}\right]-\left[O{H}^{-}\right]}{C_b}}$

Other formulas need to be used

$\begin{array}{rcl}\left[H^{+}\right]& =& {10}^{-pH}\\ \left[O{H}^{-}\right]& =& \frac{K_w}{\left[H^{+}\right]}\\ \left[{伪}_{H{A}^{-}}\right]& =& \frac{\left[H^{+}\right]K_1}{{\left[H^{+}\right]}^2+\left[H^{+}\right]K_1+K_1{K}_2}\\ \left[{伪}_{A^{2-}}\right]& =& \frac{K_1{K}_2}{{\left[H^{+}\right]}^2+\left[H^{+}\right]K_1+K_1{K}_2}\\ & & \end{array}$

Spreadsheet for differentpK₁=4.00, pK₂=8.00

From the above spreadsheet the following curve was plotted

50.0 mL of 0.020 0 M H₂A needs to be titrated with 0.100 M NaOH

. (pK₁=4.00, pK₂=6.00)

Fromtable 11.5 the equation for the fraction of titration of diprotic acid (H₂A) with strong base (B) is as follows

Other formulas need to be used

Spreadsheet forpK₁=4.00, pK₂=6.00

50.0 mL of 0.020 0 M H₂A needs to be titrated with 0.100 M NaOH

. (pK₁=4.00, pK₂=5.00)

Fromtable 11.5 the equation for the fraction of titration of diprotic acid (H₂A) with strong base (B) is as follows

Other formulas need to be used

Spreadsheet for different pK₁=4.00, pK₂=5.00

From the above spreadsheet the following curve was plotted

The family of curves will be