(Solved): Calculate the pH of a solution prepared by mixing 0.0740mol of chloroacetic acid plus 0.0240 mol ...

Calculate the pH of a solution prepared by mixing $$0.0740\mathrm{mol}$$ of chloroacetic acid plus 0.0240 mol of sodium chloroacetate in $$1.00\mathrm{L}$$ of water. First do the calculation by assuming that the concentrations of $$\mathrm{HA}$$ and $${\mathrm{A}}^{?}$$equal their formal concentrations. The $$\mathrm{p}{K}_{\mathrm{a}}$$ of chloroacetic acid is 2.865 . Then do the calculation with [HA] and $$\left[{\mathrm{A}}^{?}\right]$$from Equations 9-21 and 9-22. $$\begin{array}{l}[\mathrm{HA}]={\mathrm{F}}_{\mathrm{HA}}?\left[{\mathrm{H}}^{+}\right]+\left[{\mathrm{OH}}^{?}\right]\\ \left[{\mathrm{A}}^{?}\right]={\mathrm{F}}_{{\mathrm{A}}^{?}}+\left[{\mathrm{H}}^{+}\right]?\left[{\mathrm{OH}}^{?}\right]\end{array}$$ Using first your head, and then the Henderson-Hasselbalch equation, find the $$\mathrm{pH}$$ of a solution prepared by dissolving all the $$0.060\mathrm{mol}{\mathrm{HNO}}_3$$, and $$0.060\mathrm{mol}\mathrm{Ca}(\mathrm{OH}{)}_2,\mathrm{Ca}(\mathrm{OH}{)}_2$$ provides $$2{\mathrm{OH}}^{?}$$.