Note that the concentrations of the solutions shown on the left are millimolar (mM). This means there would be 6.023 x 1020 molecules of ONP if you had a liter of the stock solution
Todetermine the volume of 2.5 mM ONP stock solution to use to create the dilutions shown on the left, employ the following formula:
|Cs = concentration of the stock solution
Vs = volume of the stock solution
|Cf = concentration of the final solution
Vf = volume of the final solution
Begin by making a 0.225 mM ONP solution. Assume your final solution volume (Vf ) will be 10 ml. How much 2.5 mM ONP stock is needed? How much buffer?
1. Solve the equationf for Vs . Click here to check you formula.
2. Plug in the numerical values for the other variables. Click here to check your calculation.
3. Determine how much buffer to add to get a final volume of 10 ml. Click here to check your calculation.
To test your understanding of these calculations try calculating the amount of pH 7.7 buffer and 2.5 mM ONP stock solution to use to make the 0.175, 0.124, 0.075, and 0.025 mM solutions. Click here to test you calculations.
Back to making standard curves.