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Question

1. Determine the volume of 1.0M tartaric acid, H2C4H2O6, solution that when diluted to a total volume of 250.0 ml produces a solution that is 0.050M in tartaric acid
2. Determine the mass of potassium hydrogen tartrate, KHC4H4O6, required to prepare 250.0ml of a solution that is 0.050M in hydrogen tartrate ion.

Answer 1

Hello Divine

1) M1V1 = M2V2

1.0M x V1 = 0.050M x 0.2500 L

Solve for V1

V1 = 0.050M x 0.2500 L / 1.0M = 0.013 L (rounded to 2 significant figures)

If you require the volume in mL, then V1 = 0.013 L x (1000 mL / 1 L) = 13 mL

2) Potassium hydrogen tartrate is the same as hydrogen tartrate ion. The potassium ion is the counter positive cation to make the salt neutral.

mol KHC4H4O6 = 0.050 mol/L x 0.2500 L = 0.013 mol KHC4H4O6

g KHC4H4O6 = 0.013 mol KHC4H4O6 x (188.18 g KHC4H4O6 / 1 mol KHC4H4O6) = 2.4 g KHC4H4O6