(a) See answer in Lehninger on page AS–2.
(b) In the solution, we know that the total concentration of glycine (0.1 M) is equal to the concentration of glycine in the protoned form (-NH3+, the weak acid) plus the the concentration of glycine in the deprotonated form (-NH2, the conjugate base), mathematically,
$$0.1 M = [-NH_3^+] + [-NH_2].$$
This equation can be solved for the concentration of the deprotonated form
$$[-NH_2] = 0.1 M - [-NH_3^+].\quad\quad(1)$$
The Henderson-Hasselbalch equation relates the ratio of conjugate base (-NH2) to weak acid (-NH3+) to the pH of the solution and the pK value of the ionizable group
$$pH = pK + \log{{[-NH_2]} \over {[-NH_3^+]}}\quad\quad(2)$$.
Substituting (1) into (2) produces an equation with just one unkown concentration, [-NH3+],
$$pH = pK + \log{{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}.$$
Substituting the numerical values for pH and pK and solving this equation yields the [-NH3+] in this solution
$$9.0 = 9.6 + \log{{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$-0.6 = \log{{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$10^{-0.6} = {{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$0.25 = {{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$0.25[-NH_3^+] = 0.1 M - [-NH_3^+]$$
$$1.25[-NH_3^+] = 0.1 M$$
$${[-NH_3^+] = 0.08 M}.$$
Thus, the fraction of molecules in the -NH3+ form is
$${{0.08 M \over 0.1 M} = {\bf 4 \over 5}}.$$
(c) This problem is a simple titration. In order to know how much base is needed, we must know how much glycine is titrated as the solution is moved from pH 9 to 10. To do this we need to calculate the [-NH3+] at both pH 9 and 10. The difference between these two values is the [-NH3+] titrated. We'll begin by writing an expression for the [-NH2] so that we can work in terms of the [-NH3+] of interest
$$0.1M = [-NH_3^+] + [-NH_2]$$
$$[-NH_2] = 0.1 M - [-NH_3^+].$$
Now we can calculate the [-NH3+] concentration at both pH 9 and pH 10. At pH 9,
$$pH = pK + \log{{[-NH_2]} \over {[-NH_3^+]}}$$
$$pH = pK + \log{{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$9.0 = 9.6 + \log{{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$-0.6 = \log{{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$10^{-0.6} = {{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$0.25 = {{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$0.25[-NH_3^+] = 0.1 M - [-NH_3^+]$$
$$1.25[-NH_3^+] = 0.1 M$$
$$[-NH_3^+] = 0.08 M.$$
At pH 10, the calculation is similar
$$pH = pK + \log{{[-NH_2]} \over {[-NH_3^+]}}$$
$$pH = pK + \log{{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$10.0 = 9.6 + \log{{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$0.4 = \log{{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$10^{0.4} = {{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$2.5 = {{0.1 M - [-NH_3^+]} \over {[-NH_3^+]}}$$
$$2.5[-NH_3^+] = 0.1 M - [-NH_3^+]$$
$$3.5[-NH_3^+] = 0.1 M$$
$$[-NH_3^+] = 0.03 M.$$
The concentration of [-NH3+] titrated over this pH range is then
$$[-NH_3^+] = 0.08 M - 0.03 M = 0.05 M.$$
In order to calculate the volume of 5 M KOH required to titrate this concentration of -NH3+ in a volume of 1.0 L, we can use dimensional analysis. The stoichiometry is 1 mole of -NH3+ is titrated for each mole of KOH that is add (KOH is monobasic). The dimensional analysis is
$${{{0.05\ moles\ -NH_3^+} \over {1\ L}} \times {{1.0\ L} \over {}} \times {{1\ mole\ KOH} \over {1\ mol\ -NH_3^+}} \times {{1\ L\ KOH} \over {5\ moles\ KOH}} \times {{1000\ mL} \over {1\ L}} = {\bf 10\ mL}}.$$
(d) See answer in Lehninger on page AS–2.
