There is information lacking in the question. Either ýou have to specify which application you want to use the buffer for so I can make an estimate of the buffer capacity you need or, you have to specify directly which ionic strength you need.
Phosphate buffer pH 6.8 preparation protocol below Stock solutions: 0.2M dibasic sodium phosphate (1 liter) Na2HPO4*12H2O (MW=358.14) --------71.64gm + dH2O to make 1 liter (Solution X) 0.2M monobasic sodium phosphate (1 liter) NaH2PO4*H2O (MW=138.01) --------27.6gm + dH2O to make 1liter (Solution Y) Working buffer: 0.1M (1 liter) pH 6.8 245 ml solution X + 255 ml solution Y ( filled up to 1 liter with dH2O)
75g
The needed mass is 35,549 g.
1.17 grams :)
You would need 18 grams to get the requirement. Using a chemical equation can help to solve this for you.
Charles Owens got expelled from stc
The gram atomic mass of sodium is 22.9898, the formula of the least hydrated form of sodium phosphate is Na3PO4.10 H2O, and the gram formula unit mass of this sodium phosphate is 344.09. Therefore, the mass fraction of sodium in this sodium phosphate is 3(22.9898)/344.09 or about 0.20044, and the grams of sodium in 7.2 grams of this sodium phosphate is 1.44 grams of sodium, to the justified number of significant digits.
Balanced equation first. 2Na + Cl2 -> 2NaCl 35 grams NaCl (1 mole NaCl/58.44 grams )( 1 mole Cl2/2 mole NaCl )( 70.9 grams Cl2/1 mole Cl2) = 21 grams of Cl2 needed
The answer is 6,71 g dried KCl.
600 mL of 0,9 % sodium chloride: 6 x 0,9 = 5,4 grams NaCl
671
160 g (solution) - [5/100*160] g (solute) = 152 g (solvent) water