A workbook question asks:
$\pu{2.184g}$ of a solid mixture containing only $\ce{K2CO3}$ ($\mathrm{FW = 138.2058}\pu{ g/mol}$) and $\ce{KHCO3}$ ($\mathrm{FW = 100.1154} \pu{g/mol}$) is dissolved in distilled water. $\pu{31.29mL}$ of a $\pu{0.742\!M}\ \ce{HCl}$ standard solution is required to titrate the mixture to a bromocresol green end point. Calculate the weight percent of $\ce{K2CO3}$ and $\ce{KHCO3}$ in the mixture.
I started this problem by finding the amount of moles of $\ce{HCl}$ added to the titration, assuming that the solution was completely neutralized at the endpoint.
In $\pu{0.03129L}$ of a solution of $\pu{0.742\!M}\ \ce{HCl}$, there are $$\mathrm{0.03129\; \pu L \times \frac{0.742\; mol\; \ce{HCl}}{\pu L} = 0.0232\; mol\; \ce{HCl}}$$
Then I set up the balanced equations:
$$\ce{2HCl_{(aq)} + K2CO3_{(aq)} -> 2KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$ $$\ce{HCl_{(aq)} + KHCO_{3(aq)} -> KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$
This should indicate that the $\ce{HCl}$ will combine with the carbonate and bicarbonate in a 2:1 ratio, favoring the carbonate, since two moles of HCl are required to neutralize one mole of carbonate.
Then:
$$\ce{\pu{0.0232mol}\ HCl \times \frac{1\; mol\; CO_3}{2\; mol\; HCl} = 0.0116\; mol\; CO3}$$
$$\ce{0.0232\; mol\; HCl \times \frac{1\; mol\; HCO_3}{1 \pu{mol} HCl} = 0.0232\; mol\; HCO3} $$
And:
$$\ce{0.0116\; mol\; CO3 \times \frac{1\; mol\; K2CO3}{1\; mol\; CO3} = 0.0116\; mol\; K2CO3}$$
$$\ce{0.0232\; mol\; HCO3 \times \frac{1\; mol\; KHCO3}{1\; mol\; HCO3} = \pu{0.0232\; mol}\; KHCO3}$$
Converting moles to grams:
$$ \ce{0.0116\; mol\; K2CO3 \times \frac{138.2058\; g}{mol\; K2CO3} = 1.60\; g} \\ \ce{0.0232\; mol\; KHCO3 \times \frac{100.1154\; g}{mol\; KHCO3} = 2.32\; g} $$
When I add these masses together, I get $\pu{3.92 g}$, which is greater than $\pu{2.184 g}$.
Where did I go wrong?