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# 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?

Yes, your calculations are correct.

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.