ATP has $\mathrm{p} K_{\mathrm{a}}$ values of $0.9,1.5,2.3,$ and $7.7 ;$ ADP has $\mathrm{p} K_{\mathrm{a}}$ values of $0.9,2.8,$ and $6.8 ;$ and phosphoric acid has $\mathrm{p} K_{\mathrm{a}}$ values of $1.9,6.7,$ and 12.4 Do the calculation showing that at pH 7.3
a. the charge on ATP is $-3.3 .$
b. the charge on ADP is -2.8
c. the charge on phosphate is -1.8
Solution to 2a Because $\mathrm{pH} 7.3$ is much greater than the $\mathrm{p} K_{\mathrm{a}}$ values of the first three ionizations of ATP, we know that these three groups will be entirely in their basic forms at that pH, giving ATP three negative charges (Section 1.25 ). We need to determine what fraction of the group with a $\mathrm{p} K_{\mathrm{a}}$ value of 7.7 will be in its basic form at pH $7.3 .$
Because this equation has two unknowns. one of the unknowns must be expressed in terms of the other unknown. Using the definition of the acid dissociation constant $\left(K_{\mathrm{a}}\right),$ we can define [HA] in terms of $\left[\mathrm{A}^{-}\right], K_{\mathrm{a}},$ and $\left[\mathrm{H}^{+}\right]$ in order to have only one unknown.
Now we can calculate the fraction of the group with a $\mathrm{p} K_{\mathrm{a}}$ value of 7.7 that will be in the basic form. (Note that $K_{\mathrm{a}}$ is calculated from $\mathrm{p} K_{\mathrm{a}}$ and that $\left[\mathrm{H}^{+}\right]$ is calculated from pH.)
$$\frac{K_{\mathrm{a}}}{K_{\mathrm{a}}+\left[\mathrm{H}^{+}\right]}=\frac{2.0 \times 10^{-8}}{2.0 \times 10^{-8}+5.0 \times 10^{-8}}=0.3$$
Thus, the total negative charge on $\mathrm{ATP}=3.0+0.3=3.3$