Methyl orange is made by diazotising sulphanilic acid, C₆H₇NO₃S (173 g/mol), and coupling the salt…

Methyl orange is made by diazotising sulphanilic acid, $\text{C}_6\text{H}_7\text{NO}_3\text{S}$ ($173\ \text{g/mol}$), and coupling the salt with N,N-dimethylaniline, $\text{C}_8\text{H}_{11}\text{N}$ ($121\ \text{g/mol}$). The sequence is a standard undergraduate preparation of a laboratory indicator. Assume that the sodium nitrite and the acid are used in exactly the amounts needed. The product is the sodium salt of the dye, $\text{C}_{14}\text{H}_{14}\text{N}_3\text{NaO}_3\text{S}$ ($327\ \text{g/mol}$). A student starts with $17.3\ \text{g}$ of sulphanilic acid. The diazotisation uses sodium nitrite ($69\ \text{g/mol}$), and the whole sequence has a yield of $60\ \%$. Methyl orange is also an indicator, with $\text{p}K_a = 3.47$ for the change from the red acid form to the yellow base form. Find the mass of sodium nitrite needed, the mass of N,N-dimethylaniline needed, the most dye that could form, and the mass collected. Then find the ratio of the yellow form to the red form at pH 4.40.