J.R. S. answered 10/01/18
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Ph.D. in Biochemistry--University Professor--Chemistry Tutor
a. (X ml)(1.0 M) = (250.0 ml)(0.05 M)
x = 12.5 mls needed
b. Molar mass potassium hydrogen tartrate = 188.177g/mol. To prepare 0.05 M hydrogen tatrate is the same as preparing 0.05 M potassium hydrogen tartrate because KHC4H4O6 => K+ + HC4H4O6-
0.05 mol/L x 0.25 L x 188.2 g/mol = 2.35 g needed