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QUESTION

# HIll buffer is a solution of 0.4 M K2HPO4, made up in .08M KCl. We need 500 mL for the week. How many grams of K2HPO4 (MW=174.17g) were need to make this volume of buffer? My wkr: 1mole/174.1g x 1L/.4moles = 1L/69.67g 69.67g/1L x.5L = 34.83g Rightorwrong?

You can use the of the solution to determine how many moles of dipotassium phosphate you'd get in 500 mL, then multiply that value by the molar mass of the compound

C = n/V => n = C * V

n_(K_2HPO_4) = 0.4"moles"/cancel("L") * 500 * 10^(-3)cancel("L") = "0.2 moles " K_2HPO_4

and

0.2cancel("moles "K_2HPO_4) * "174.17 g"/(1cancel("mole "K_2HPO_4)) = "34.83 g " K_2HPO_4

If you take into account , the answer should be

m_(K_2SO_4) = "30 g" -> since you only give one sig fig for the volume and of the dipotassium phosphate.