Question 15.9: Design of a Glass We produce good chemical resistance in a g......
Design of a Glass
We produce good chemical resistance in a glass when we introduce B_{2}O_{3} into silica. To ensure that we have good glass-forming tendencies, we wish the O:Si ratio to be no more than 2.5, but we also want the glassware to have a low melting temperature to make the glass-forming process easier and more economical. Design such a glass.
Because B_{2}O_{3} reduces the melting temperature of silica, we would like to add as much as possible. We also, however, want to ensure that the O:Si ratio is no more than 2.5, so the amount of B_{2}O_{3} is limited. As an example, let us determine the amount of B_{2}O_{3} we must add to obtain an O:Si ratio of exactly 2.5. Let f_{B} be the mole fraction of B_{2}O_{3} added to the glass, and (1 – f_{B}) be the mole fraction of SiO_{2}:
\frac{O}{Si} =\frac{(3 \ \frac{O \ ions}{B_{2}O_{3}}) ( f_{B}) \ + \ (2 \ \frac{O \ ions}{SiO_{2}}) (1 \ – \ f_{B})}{(1 \ \frac{Si \ ion}{SiO_{2}})(1 \ – \ f_{B})}= 2.5
3f_{B} \ + \ 2 \ – \ 2f_{B} = 2.5 \ – \ 2.5f_{B} or f_{B} = 0.143
Therefore, we must produce a glass containing no more than 14.3 mol% B_{2}O_{3}. In weight percent:
wt% B_{2}O_{3} = \frac{( f_{B})(69.62 \ g/mol)}{( f_{B})(69.62 \ g/mol) \ + \ (1 \ – \ f_{B})(60.08 \ g/mol)}× 100
wt% B_{2}O_{3} = \frac{(0.143)(69.62 \ g/mol)}{(0.143)(69.62 \ g/mol) \ + \ (0.857)(60.08 \ g/mol)} × 100 = 16.2