because that tip of the calomel electrode is made up of porous nature..If we keep it in environment, its get oxidised,,
I just got done talking with a vendor for pH probes and he explained that the KCl solution helps facilitate the measurement of the hydrogen ions with the electrode. It will actually seep out in small quantities from the tip which is a membrane.
No. Potassium chloride (KCl) is soluble in water.
Molar mass of KCl = 74.55g/mol.ie, if you dissolve 74.55g KCl in 1litre (1000 ml) of water, it will be 1M KCl solution.If you want to make 3M KCl solution,Dissolve 3 ×74.55 = 223.65g KCl in 1litre (1000 ml) of water.If you want to make different molar solutions of KCl, just calculate as per below given equation.Weight of KCl to be weighed =Molarity of the solution needed × Molecular weight of KCl (ie, 74.55) × Volume of solution needed in ml / 1000.To prepare 3M KCl in 1 litre, it can be calculated as follows,3 mol × 74.55 g/mol × 1000 ml / 1000 ml = 223.65gByPraveen P Thalichalam, Kasaragod (Dist), Kerala.
The initial amount of KCl in the solution is 12mL (12% of 100mL). When 200mL of water is added, the total volume becomes 300mL. You would then divide the initial amount of KCl (12mL) by the total volume (300mL) and multiply by 100 to get the final concentration of KCl in the solution.
I believe it should be 7.0. Here's an example of 3M KCl made for sale with the specs, showing a pH of 7.0
It consists of tube in the bottom of which is a layer of mercury ,over which is placed a paste of Hg+HgCl2.The remaining portion of cell is filled with a solution of normal or decinormal or saturated solution of KCl. a platinum wire dipping into the mercury layer is used for making electrical contact. the side tube is used for electrode is formulated as;Hg,Hg2Cl2,KCl. the electrode can be coupled with the hydrogen electrode of unknown pH.
moles KCl = ( M solution ) ( V solution in L )moles KCl = ( 2.2 mol KCl / L solution ) ( 0.635 L of solution )moles KCl = 1.397 moles KCl
I just got done talking with a vendor for pH probes and he explained that the KCl solution helps facilitate the measurement of the hydrogen ions with the electrode. It will actually seep out in small quantities from the tip which is a membrane.
A potassium chloride (KCl) solution is colorless.
moles KCL = ( M solution ) ( L of solution )moles KCl = ( 0.83 mol KCl / L ) ( 1.7 L ) = 1.41 moles KCl
To prepare a 500mM KCl solution, you would need to dissolve 74.55 grams of KCl in enough solvent to make 1 liter of solution.
M= moles in solution/liters so plug in what you know 3.0M of KCl solution = moles in solution/ 2.0L multiply both sides by 2.0L moles solute = 1.5 moles KCl so you need 1.5 moles KCl to prepare the solution
To find the volume of the solution, first calculate the number of moles of KCl in 2.39g using its molar mass (74.55 g/mol). Then, use the molarity equation (Molarity = moles of solute / volume of solution in liters) to find the volume of the solution.
Need mole KCl first. 4.88 grams KCl (1 mole KCl/74.55 grams) = 0.06546 moles KCl =======================now, Molarity = moles of solute/Liters of solution ( 423 ml = 0.423 Liters ) Molarity = 0.06546 moles KCl/0.423 Liters = 0.155 M KCl ------------------
I did not know that you could get a concentration of 75.66 M KCl, but; Molarity = moles of solute/Liters of solution 75.66 M KCl = moles KCl/1 liter = 75.66 moles of KCl 75.66 moles KCl (74.55 grams/1 mole KCl) = 5640 grams KCl that is about 13 pounds of KCl in 1 liter of solution. This is why I think there is something really wrong with this problem!
The remaining 10 g of KCl forms a saturated solution in water, meaning the solution is in dynamic equilibrium with solid KCl dissolving and KCl ions re-crystallizing at the same rate. The resulting solution will contain both dissolved K+ and Cl- ions in equilibrium with undissolved KCl crystals.
This is a solution of 10 g KCl/100 g water.