I'm going to assume you want 100ppm by mass, since this is really the only way that makes sense with a solid and liquid.
First, you need to get a fluoride salt. I will use sodium fluoride as an example - it may or may not matter which salt depending on what you're trying to do. Be careful, fluoride compounds can be toxic.
Second, you need to figure out what percentage of the salt is actually fluoride. To do this in our example, first look up the molecular weights of sodium and fluorine on a Periodic Table (the numbers at the bottom of the squares). Sodium (Na) has a molecular weight of 22.989, Fluorine (F) weighs 18.998. Add these to get the total molecular weight of sodium fluoride:
22.989 + 18.998 = 41.987
Now, divide the total weight by the weight of fluorine to get a conversion factor (units of NaF / units of F):
41.987 / 18.998 = 2.21
Third, you need to weigh out your fluorine and water - 100 parts fluorine for every (million - 100) parts water. This reduces to 1 part fluorine per 9,999 parts water (I'll round to 10,000). This is a big difference, so you may need two different scales with different ranges, and it's going to make a bunch of solution.
I'm going to measure in grams, but you could use any measure of weight. First, weight out 10,000 g (10 kg) of water and set it aside. Remember to subtract out the weight of the container when weighing! Next, we need 1 g of fluoride. However, we have sodium fluoride, not fluorine, so we need to use the conversion factor from before:
1 g (F) * 2.21 (NaF / F) = 2.21 g NaF
Weigh out 2.21 g of your salt, add it to the water, and you're done!!!
A problem with this method is you'll end up with a huge amount of solution, and it may be more water than you can easily deal with. To avoid the issue, make the initial solution more concentrated, then dilute it again in a second step. For example, use only 1 / 10th the amount of water, and you'll get a 1,000 ppm solution. You then take 1 / 10th of that solution and add it to 9/10 water in another container. This will give you a 100 ppm solution and you don't have to deal with such a large amount of liquid. You can even add another step for very dilute solutions.
Fluoride is a weak base, not an acid. When fluoride ions are in solution, they can accept protons to form HF, which is a weak acid.
It has something to do with pharmaceutical calculations and ratios for mixing drugs within other solubles. In this case, 0.14% w/v fluoride ion means it contains .14% "weight by volume" of fluoride in ratio to the other ingredients. There are other abbreviations that you may have seen on other common things in your home such w/w or v/v (weight by weight and volume by volume, respectively).
The fluoride ion is noted as F- (F minus).
The usual ion form of fluorine is the fluoride ion, which has a charge of -1.
When potassium fluoride is dissolved in water, it dissociates into its ions: K+ and F-. This creates a solution that is basic in nature due to the presence of the fluoride ion. The fluoride ion can also form complexes with other ions present in the solution.
The Nernstian response of a fluoride ion-selective electrode refers to its ability to produce a theoretical slope of -59.2 mV per decade change in fluoride ion concentration, under ideal conditions. This response is based on the Nernst equation and reflects the electrode's sensitivity to changes in fluoride ion concentration in a solution.
No, The fluoride ion is a reduced form of fluorine.
The charge of a fluoride ion is -1, as it gains one electron to achieve a stable electron configuration following the octet rule.
The chemical formula of a sodium ion is Na+, while the chemical formula of a fluoride ion is F-.
Fluoride ion (F-) is not an element, but an ion formed from the element fluorine (F). Fluorine is a chemical element on the periodic table, while fluoride ion is a negatively charged particle resulting from the gain of an electron by a fluorine atom.
Fluoride Ion
When a lithium ion is attracted to a fluoride ion, they may form an ionic bond to create lithium fluoride. The positively charged lithium ion is attracted to the negatively charged fluoride ion due to their opposite charges. This bond is typically strong and stable.