Question:
The enamel present on teeth becomes much harder due to the conversion of hydroxyapatite into fluorapatite. What are \( X \) and \( Y \)?
The enamel present on teeth becomes much harder due to the conversion of hydroxyapatite into fluorapatite. What are \( X \) and \( Y \)?
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Fluoride strengthens enamel by converting hydroxyapatite into fluorapatite.
Updated On: May 18, 2025
- \( X = \text{Ca(OH)}_2, Y = \text{CaF}_2 \)
- \( X = \text{Ca(OH)}_2, Y = \text{CaCl}_2 \)
- \( X = \text{Ca(OH)}_2, Y = \text{NaCl} \)
- \( X = \text{CaO}, Y = \text{CaCl}_2 \)
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The Correct Option is A
Approach Solution - 1
Step 1: Understanding Tooth Enamel Hardening
- Hydroxyapatite: \( \text{Ca}_5(\text{PO}_4)_3\text{OH} \)
- Fluoride treatment replaces OH⁻ with F⁻, forming fluorapatite, which is stronger.
\[ 3\text{Ca}_5(\text{PO}_4)_3\text{OH} + F^- \rightarrow 3\text{Ca}_5(\text{PO}_4)_3\text{F} \]
- \( X = \text{Ca(OH)}_2 \) contributes OH⁻.
- \( Y = \text{CaF}_2 \) provides F⁻.
- Hydroxyapatite: \( \text{Ca}_5(\text{PO}_4)_3\text{OH} \)
- Fluoride treatment replaces OH⁻ with F⁻, forming fluorapatite, which is stronger.
\[ 3\text{Ca}_5(\text{PO}_4)_3\text{OH} + F^- \rightarrow 3\text{Ca}_5(\text{PO}_4)_3\text{F} \]
- \( X = \text{Ca(OH)}_2 \) contributes OH⁻.
- \( Y = \text{CaF}_2 \) provides F⁻.
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Approach Solution -2
The enamel on teeth mainly consists of hydroxyapatite, which has the chemical formula:
\[ \text{Ca}_{10}(\text{PO}_4)_6(\text{OH})_2 \]
Hydroxyapatite can be converted into fluorapatite, which is harder and more resistant to acid attack. This conversion improves the strength and durability of tooth enamel and helps prevent dental caries.
The conversion reaction involves two key substances:
- \( X = \text{Ca(OH)}_2 \) (calcium hydroxide), which provides hydroxide ions.
- \( Y = \text{CaF}_2 \) (calcium fluoride), which supplies fluoride ions for the substitution.
In the presence of fluoride ions from calcium fluoride, the hydroxide ions (OH⁻) in hydroxyapatite are replaced by fluoride ions (F⁻), forming fluorapatite:
\[ \text{Ca}_{10}(\text{PO}_4)_6(\text{OH})_2 + 2 F^- \rightarrow \text{Ca}_{10}(\text{PO}_4)_6\text{F}_2 + 2 OH^- \]
Thus, calcium hydroxide and calcium fluoride are essential for the formation of fluorapatite, which strengthens tooth enamel.
Therefore, the correct values are:
\[ X = \text{Ca(OH)}_2, \quad Y = \text{CaF}_2 \]
\[ \text{Ca}_{10}(\text{PO}_4)_6(\text{OH})_2 \]
Hydroxyapatite can be converted into fluorapatite, which is harder and more resistant to acid attack. This conversion improves the strength and durability of tooth enamel and helps prevent dental caries.
The conversion reaction involves two key substances:
- \( X = \text{Ca(OH)}_2 \) (calcium hydroxide), which provides hydroxide ions.
- \( Y = \text{CaF}_2 \) (calcium fluoride), which supplies fluoride ions for the substitution.
In the presence of fluoride ions from calcium fluoride, the hydroxide ions (OH⁻) in hydroxyapatite are replaced by fluoride ions (F⁻), forming fluorapatite:
\[ \text{Ca}_{10}(\text{PO}_4)_6(\text{OH})_2 + 2 F^- \rightarrow \text{Ca}_{10}(\text{PO}_4)_6\text{F}_2 + 2 OH^- \]
Thus, calcium hydroxide and calcium fluoride are essential for the formation of fluorapatite, which strengthens tooth enamel.
Therefore, the correct values are:
\[ X = \text{Ca(OH)}_2, \quad Y = \text{CaF}_2 \]
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