(Solved): The Determination of Fluoride Using an Ion Selective Electrode: The following data was collected dur ...

Four concentration of fluoride standards will be prepared $$(0.1,1.0,5.0$$ and $$10.0\mathrm{ppm}$$ fluoride ion) will be prepared from a stock solution of $$100\mathrm{ppm}$$. Calculate the volume needed to prepare $$50\mathrm{mL}$$ of each solution with different concentration. To each of this solution, $$25\mathrm{mL}$$ of TISAB buffer will be added and the total volume will be made to $$50\mathrm{mL}$$ with deionized (DI) water. For the water sample, $$25\mathrm{mL}$$ of tap water and diluted to $$50\mathrm{ml}$$ with the addition of $$25\mathrm{mL}$$ of TISAB. The dilution factor is 2 (25 to 50 ). Take the ppm F obtained from the graph and multiply it by 2. For the mouth wash samples, take $$2\mathrm{mL}$$ of sample, and add it with $$25\mathrm{mL}$$ of TISAB, and dilute it to $$50\mathrm{mL}$$. The dilution factor is $$25(2$$ to 50 ). Take $$\mathrm{ppm}\mathrm{F}$$ from the graph and multiply it by 25 to report the final ppm $$\mathrm{F}$$. If you divide this by 10,000 , you get the $$\%\mathrm{F}$$ in mouth wash, which should be around $$0.02\%$$. For the toothpaste sample, diluted it to $$100\mathrm{mL}$$ using a dilution factor of 100 . Multiply the ppm $$\mathrm{F}$$ from the graph by 100 , and then as above, divide by 10,000 to get the $$\%\mathrm{F}$$. The tube indicates $$0.16\%$$ fluoride. A blank consisting of $$25\mathrm{mL}$$ of TISAB and $$25\mathrm{mL}$$ of deionized water will also be prepared. The electrode will immersed into the solution, and when it stabilized (usually after ca. $$3\min$$ ), the readings will be recorded. After each sample reading, rinse the electrode with deionized water to prevent any solution carry-over.