It is a solution whose strength and quantity are given .
take 7.5 gram of sodium carbonate and dissovled in 100 ml of water..thats it.. this solution is called 7.5% sodium carbonate soultion.
I did this science practical at school i added half a teaspoon of sodium carbonate to 20 ml of vinegar. It fizzed up then after about ten seconds went down.
use the equation that is standard: 1000 ml 1 M solution= (MOLECULAR WEIGHT) X ml 0.05 M solution = ((MOLECULAR WEIGHT)*X*0.05)/1000
Molarity is moles per litre. You need to divide the moles by 1500 and multiply by 1000. 5.6/1500*1000 is 3.73333 molar.
Dissolve 50 g of potassium carbonate in 100 mL of water at 20 0C.
take 7.5 gram of sodium carbonate and dissovled in 100 ml of water..thats it.. this solution is called 7.5% sodium carbonate soultion.
Firstly, they'll react each other forming sodium hydrogen carbonate and sodium chloride. If there is excess HCl, the sodium hydrogen carbonate would further react till sodium chloride and evolve carbon dioxide.
1) 0.10 mol of solid sodium hydrogen carbonate and 0.20 mol of solid sodium carbonate are dissolved in the same beaker of water, transferred to a volumetric flask and made to 250.0 mL. The Ka for HCO3 - is 4.7 x 10-11. a) What is the pH of the resulting buffer? b) What is the pH of solution after 20.00 mL of 0.050 mol L-1 hydrochloric acid solution is added to 25.00 mL of the original solution? c)What is the pH of the resulting buffer after 0.040 g of solid sodium hydroxide is added to 25.00 mL of the original solution? 2) Plan how you would make 100.0 mL of a buffer solution with a pH of 10.80 to be made using only sodium carbonate, sodium hydrogen carbonate and water.Youshould specify the amounts of sodium carbonate and sodium hydrogen carbonate that you would use. ( the ration acid to base is 3:1)
I did this science practical at school i added half a teaspoon of sodium carbonate to 20 ml of vinegar. It fizzed up then after about ten seconds went down.
The density of sodium carbonate is about 2.54 grams per milliliter. Therefore, 10 mL of sodium carbonate would weigh approximately 25.4 grams.
Molarity = moles of solute/Liters of solution ( 1500 mL = 1.5 Liters ) Molarity = 0.800 moles NaOH/1.5 Liters = 0.533 M sodium hydroxide ...
Mix 100 mL of a 1 N solution with 900 mL of distilled water.
First, calculate the moles of sodium ions in each solution. Then, determine the total volume of the combined solutions. Finally, divide the total moles of sodium ions by the total volume in liters to get the concentration of sodium ions.
Molarity = moles of solute/Liters of solution ( 1500 mL = 1.5 Liters ) Molarity = 0.800 moles NaOH/1.5 Liters = 0.533 M sodium hydroxide ...
To prepare a 100 mM Na2CO3/NaHCO3 buffer at pH 9, you will need to mix Na2CO3 and NaHCO3 in specific proportions to achieve the desired pH. You can calculate the exact ratio using the Henderson-Hasselbalch equation and adjust the ratio of the two components based on the desired pH. Then, dissolve the calculated amounts of Na2CO3 and NaHCO3 in water, adjust the pH with either HCl or NaOH, and dilute to the final volume.
To precipitate magnesium ion, you can add a precipitating agent such as sodium hydroxide (NaOH) to the magnesium nitrate solution. This will cause magnesium hydroxide (Mg(OH)2) to form as a precipitate. You can then filter the solution to separate the precipitate from the liquid.
use the equation that is standard: 1000 ml 1 M solution= (MOLECULAR WEIGHT) X ml 0.05 M solution = ((MOLECULAR WEIGHT)*X*0.05)/1000