To standardize 0.02 N NaOH, you would titrate it against a known concentration of an acid, such as a primary standard like potassium hydrogen phthalate. By carefully adding the NaOH to the acid until the endpoint is reached, you can determine the exact concentration of the NaOH solution.
Sodium Carbonate (Na2CO3) is most commonly used to standardise acids as it is non-hygroscopic (doesn't absorb water from the atmosphere).Oxalic Acid (HOOCCOOH or H2C2O4) used to standardise bases
In order to make 0.02 N NaOH from 0.2 N NaOH, one needs to dilute it by 10 x (10 fold). Depending on the volume of 0.02 N NaOH needed, that will determine the volume of 0.2 N used. For example, to make 100 ml of 0.02 N NaOH, you would dilute 10 mls of 0.2 N to 100 ml. This is seen in the following calculation: (x ml)(0.2 N NaOH) = (100 ml) (0.02 N NaOH) and x = 10 ml
To calculate the concentration of a 0.1 N NaOH solution, you need to know the molar mass of NaOH which is approximately 40 g/mol. Since 1 N solution is equivalent to 1 gram equivalent weight of the solute dissolved in 1 liter of the solution, a 0.1 N NaOH solution would contain 4 g of NaOH per liter of solution.
To prepare a 2.5 N NaOH solution in methanol, you would first calculate the amount of NaOH needed based on its molar mass and the desired concentration. Then, dissolve this amount of NaOH in the appropriate volume of methanol to make the 2.5 N solution. Be sure to wear appropriate safety gear and follow proper handling procedures when working with NaOH.
To prepare a 10 N solution of NaOH, you would dissolve 400 g of NaOH pellets in enough distilled water to make 1 liter of solution. Remember to add the NaOH to the water slowly and carefully due to the exothermic nature of the reaction. Finally, stir the solution until all of the NaOH has fully dissolved.
Sodium Carbonate (Na2CO3) is most commonly used to standardise acids as it is non-hygroscopic (doesn't absorb water from the atmosphere).Oxalic Acid (HOOCCOOH or H2C2O4) used to standardise bases
In order to make 0.02 N NaOH from 0.2 N NaOH, one needs to dilute it by 10 x (10 fold). Depending on the volume of 0.02 N NaOH needed, that will determine the volume of 0.2 N used. For example, to make 100 ml of 0.02 N NaOH, you would dilute 10 mls of 0.2 N to 100 ml. This is seen in the following calculation: (x ml)(0.2 N NaOH) = (100 ml) (0.02 N NaOH) and x = 10 ml
To calculate the concentration of a 0.1 N NaOH solution, you need to know the molar mass of NaOH which is approximately 40 g/mol. Since 1 N solution is equivalent to 1 gram equivalent weight of the solute dissolved in 1 liter of the solution, a 0.1 N NaOH solution would contain 4 g of NaOH per liter of solution.
Weigh 20g. sodium hydroxide pillets pour it into a clean and dry 1L measuring flask. Dilute it with 200 mL of distilled water vith vigourous shaking, then dilute with distilled water up to the mark of 1000mL. Stand it for few hours. Check, it should be clear solution and then standardise it with potassium hydrogen phthallate OR oxalic acid.
To prepare a 2.5 N NaOH solution in methanol, you would first calculate the amount of NaOH needed based on its molar mass and the desired concentration. Then, dissolve this amount of NaOH in the appropriate volume of methanol to make the 2.5 N solution. Be sure to wear appropriate safety gear and follow proper handling procedures when working with NaOH.
To prepare a 10 N solution of NaOH, you would dissolve 400 g of NaOH pellets in enough distilled water to make 1 liter of solution. Remember to add the NaOH to the water slowly and carefully due to the exothermic nature of the reaction. Finally, stir the solution until all of the NaOH has fully dissolved.
formula for neutralizatrion is volume of acid X normality of acid = volume of base X normality of base so (0.3)(3) should equal (4)(volume) which is .225L. However, Ca(OH)2 contains 2 moles of OH resulting division of total volume needed by 2. Thus, the answer becomes .1125L or 112.5ml.
Facotires need to standardise manufactured parts
First, calculate the number of moles of NaOH: Moles = Molarity x Volume (L) Convert mL to L: 450 mL = 0.45 L Moles = 0.25 N x 0.45 L = 0.1125 moles of NaOH.
standardise means making everythig the same.
Since "normality" is defined as the gram equivalent weight of a substance in a liter of solution, a 0.02 N NaOH solution would have 0.02 gram equivalents of NaOH per liter. To reduce it to 0.01 N you need only dilute it to one half of the original - e.g. 500 ml of NaOH mixed with 500 ml of pure water. Because there is a small change in the density upon mixing, the exact amount of water will differ slightly from 500 ml, but for a solution as dilute as 0.02 N, it won't be that far off. The best way to get it exact would be to start with a known volume of 0.02 N NaOH and then add enough water to bring the total to exactly twice the original volume. This might be accomplished by doing it in a graduated cylinder or adding it from a burette into a volumetric flask. the important thing is to know the starting volume of the 0.02 N solution and the final volume of the diluted (0.01 N) solution
It cannot N is Nitrogen where as Na is Sodium, they are two totally different elements