A student plans to measure the concentration of a basic sample by titration with HCℓ. In...
A student plans to measure the concentration of a basic sample by titration with HCℓ. In order to standardize the HCℓ, a primary standard solution is made by weighing 0.7235 g of "tris" base (H2NC(CH2OH)3 - it is monobasic), quantitatively transferring the powder to a 500 mL volumetric flask, and diluting to volume with deionized water. A 10.00 mL aliquot of the tris solution is taken for titration. What is the concentration of tris in the primary standard, in M?
I got an answer of 0.011945 M but was confused with the number of significant figures that we needed to use. Would it just be 1 sig fig due to 500 mL? That almost seems incorrect. Thank you for your help.
Homework Answers
The concentration of tris is calculated:
[Tris] = g / MM * V = 0.7235 g / 121.1 g / mol * 0.5 L = 0.01195 M
4 significant figures should be used for the use of 0.7235 of the mass.
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