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sodium phosphate dibasic means that there are two sodium molecules in the formula (contributing to two bases). the formula is Na2HPO4. the first step is to figure out the molecular weight of the compount with the help of a Periodic Table. This comes out to 142g/mole. To make 1M solution (M= moles/liter so I am assuming 1 liter) you need 142 grams of the compound in 1L of dH2o. The definition of molarity (which is moles over liters) makes the calculation easy for making 1L of the solution. However if you want to make less or more of the solution, you will have to calculate accordingly. For example if you wanted to make 50 ml (0.05L) of the solution, you would multiply 142 g/liter (g/mol x moles/liter will give g/liter) X 0.05 liter, which would give you 7.1 g. 7.1 g of Na2HPO4 is needed to make 50 ml of 1 M solution. 142 g is needed to make 1000ml (1 L) of 1M solution.
What else can I help you with?
What does 1M solution of sodium carbonate mean?
A 1M solution of sodium carbonate means that it contains 1 mole of sodium carbonate dissolved in 1 liter of solvent (usually water). This concentration is used in chemistry to describe the amount of the solute (sodium carbonate) present in the solution.
How much sodium chloride is required to make 1litre of 1M solution?
To make a 1M solution of sodium chloride in 1 liter of water, you would need 58.44 grams of sodium chloride. This is based on the molecular weight of sodium chloride, which is 58.44 g/mol.
How do you prepare phosphate buffer solution of pH 8?
See the Web Links to the left of this answer.I especially like the Smith.edu link -- it has complete and very useful description of how to prepare a buffer.Use the Henderson-Hasselbach equation:pH = pKa + log [A-]/[HA]where HA is the protonated form of the weak acid, A- is the salt (dissociated acid, or in other words, its conjugate base), and the pKa is the strength of the acid.What this says is that the pH that you want your buffer to be depends on two things:-- the pKa of the weak acid you are using (see reference tables under the Web Links to the left)-- and the RATIO of the concentration of the acid and the salt that you add to the solution.The pH of the buffer does not depend on the actual concentration of the buffer, but on the ratio of the two parts.The buffer capacity depends on two things -- how close to the pKa the pH of the buffer actually is (it should be within 1-2 pH units), and what the total concentration of the buffer is.For instance if you have 0.001 M acetic acid and 0.001 M sodium acetate, the resulting buffer will have the exact same pH as a buffer made with 0.1 M acetic acid and 0.1 M sodium acetate (because the ratio is 1 to 1, the pH = pKa = 4.76). However, the 0.1 M buffer will have a much larger buffer capacity, and will much better resist changes in pH upon the addition of a strong acid or base.
How do you prepare a solution containing 300ml of 1M Sodium Sulfate?
That is a poorly worded question. The way you have worded it could be answered correctly by instructing you to "Pour 300ml of 1M solution in a swimming pool."The question is usually posed in the form of "How much sodium sulfate is needed to produce 300 ml of 1M solution?" Presuming you intend the latter, you would prepare 300 ml of 1M solution by mixing 300/1000 of a mole of sodium sulfate with sufficient water to produce 300 ml of solution.Note the difference between Molar and Molal: 1M = 1 mole per liter of solution, 1 molal = 1 mole per liter of water. This distinction has a major effect on the wording of your answer.300 ml / 1000 ml = 3/10 = 0.3000One mole of sodium sulfate (Na2SO4) has the formula weight of 142.04314 g/mol as shown below:fw Na x 2 = 22.98977 g/mol x 2 = 45.97954 g/molfw S = 32.066 g/molfw O x 4 = 15.994 g/mol x 4 = 63.9976 g/mol45.97954 g/mol + 32.066 g/mol + 63.9976 g/mol = 142.04314 g/molThe mass of 0.3000 mol x 142.04314 g/mol = 42.613 g.Note: The final answer was rounded to the thousandths place because the formula weight for Sulfur was the least precise term.
How many moles of sodium ions and chloride ions are present in 1 liter of a 1M solution of sodium chloride?
In a 1M solution of sodium chloride, there would be 1 mole of sodium ions and 1 mole of chloride ions in 1 liter of the solution. This is because each formula unit of sodium chloride dissociates into one sodium ion and one chloride ion in solution.
What does 1M solution of sodium carbonate mean?
A 1M solution of sodium carbonate means that it contains 1 mole of sodium carbonate dissolved in 1 liter of solvent (usually water). This concentration is used in chemistry to describe the amount of the solute (sodium carbonate) present in the solution.
How can we prepare 1M solution of NaCl?
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
How much sodium chloride is required to make 1litre of 1M solution?
To make a 1M solution of sodium chloride in 1 liter of water, you would need 58.44 grams of sodium chloride. This is based on the molecular weight of sodium chloride, which is 58.44 g/mol.
Which phase will be on top in an immiscible mixture of methylene chloride and 1M of sodium hydroxide?
Sodium hydroxide solution will be on the top.
How do you prepare 1M from 10mM Tris-HCL?
To prepare 1M Tris-HCl from a 10mM solution, you would need to dilute the 10mM solution by a factor of 100. This means you would mix 1 part of the 10mM solution with 99 parts of water to achieve a final concentration of 1M Tris-HCl.
How do you prepare phosphate buffer solution of pH 8?
See the Web Links to the left of this answer.I especially like the Smith.edu link -- it has complete and very useful description of how to prepare a buffer.Use the Henderson-Hasselbach equation:pH = pKa + log [A-]/[HA]where HA is the protonated form of the weak acid, A- is the salt (dissociated acid, or in other words, its conjugate base), and the pKa is the strength of the acid.What this says is that the pH that you want your buffer to be depends on two things:-- the pKa of the weak acid you are using (see reference tables under the Web Links to the left)-- and the RATIO of the concentration of the acid and the salt that you add to the solution.The pH of the buffer does not depend on the actual concentration of the buffer, but on the ratio of the two parts.The buffer capacity depends on two things -- how close to the pKa the pH of the buffer actually is (it should be within 1-2 pH units), and what the total concentration of the buffer is.For instance if you have 0.001 M acetic acid and 0.001 M sodium acetate, the resulting buffer will have the exact same pH as a buffer made with 0.1 M acetic acid and 0.1 M sodium acetate (because the ratio is 1 to 1, the pH = pKa = 4.76). However, the 0.1 M buffer will have a much larger buffer capacity, and will much better resist changes in pH upon the addition of a strong acid or base.
How do you prepare 0.02m Naoh from 1M Naoh solution?
To prepare 0.02M NaOH from 1M NaOH solution, you will need to dilute the 1M solution. Use the formula: C1V1 = C2V2, where C1 is the concentration of the stock solution (1M), V1 is the volume of the stock solution you will use, C2 is the desired concentration (0.02M), and V2 is the final volume of the diluted solution. Calculate the volume of 1M NaOH solution (V1) needed to make the desired 0.02M concentration and dilute it with water to reach the desired volume (V2).
What is produced at the cathode of an electrolytic cell containing 1M solution of sodium chloride?
Chlorine gas. Cl2
How do you prepare a solution containing 300ml of 1M Sodium Sulfate?
That is a poorly worded question. The way you have worded it could be answered correctly by instructing you to "Pour 300ml of 1M solution in a swimming pool."The question is usually posed in the form of "How much sodium sulfate is needed to produce 300 ml of 1M solution?" Presuming you intend the latter, you would prepare 300 ml of 1M solution by mixing 300/1000 of a mole of sodium sulfate with sufficient water to produce 300 ml of solution.Note the difference between Molar and Molal: 1M = 1 mole per liter of solution, 1 molal = 1 mole per liter of water. This distinction has a major effect on the wording of your answer.300 ml / 1000 ml = 3/10 = 0.3000One mole of sodium sulfate (Na2SO4) has the formula weight of 142.04314 g/mol as shown below:fw Na x 2 = 22.98977 g/mol x 2 = 45.97954 g/molfw S = 32.066 g/molfw O x 4 = 15.994 g/mol x 4 = 63.9976 g/mol45.97954 g/mol + 32.066 g/mol + 63.9976 g/mol = 142.04314 g/molThe mass of 0.3000 mol x 142.04314 g/mol = 42.613 g.Note: The final answer was rounded to the thousandths place because the formula weight for Sulfur was the least precise term.
How can you prepare 1M H2SO4?
To prepare 1M H2SO4 solution, you would need to dilute concentrated sulfuric acid (approximately 18M) by adding the appropriate amount of water. To make 1L of 1M H2SO4 solution, you would mix approximately 55.5 mL of concentrated sulfuric acid with about 944.5 mL of water in a volumetric flask while taking proper safety precautions.
How to prepare 1m hcl from 35 percent hcl?
To prepare 1M HCl solution from 35% HCl solution, you would need to dilute the 35% HCl with water. Use the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the volume of the initial solution, C2 is the final concentration (1M), and V2 is the final volume (1 liter in this case). Calculate the volume of 35% HCl needed to achieve a 1M solution, then add water to make up the total volume to 1 liter.
How many moles of sodium ions and chloride ions are present in 1 liter of a 1M solution of sodium chloride?
In a 1M solution of sodium chloride, there would be 1 mole of sodium ions and 1 mole of chloride ions in 1 liter of the solution. This is because each formula unit of sodium chloride dissociates into one sodium ion and one chloride ion in solution.
