Question

If 42.1 mL of 1.02 M sodium hydroxide, measured using a graduated cylinder, is placed in a beaker filled with 300 mL of DI water, what is the concentration of the diluted NaOH solution?

b) Why does this calculation only provide an estimate of the NaOH concentration? In other words, why do we have to standardize the NaOH in this experiment to find its exact concentration?

Answer 1

Approximately 0.126 M

Step-by-step explanation:

For the calculation of the dilution you take into account the moles of NaOH in the 42.1mL of the original solution and you use the new volume of 342.1 mL:

`(42.1 mL * 1.02 M ) Number Of Moles\\\\C=(42.1mL*1.02M)/(342.1 mL) = 0.126 M`

The standardization is necessary because a beaker is not not an instrument used to measure volumes and the marks on it only give an estimate of the volume of the solution, they are used to contain solutions and carry reactions among other things. If you would have measured the water with a graduated cylinder (an instrument designed to measure volumes) the standardization wouldnt be that necessary.