Question

A student was asked to standardize a solution of NaOH to four significant figures. The student used a primary standard acid (MW = 204.22 g/mol), dissolved in DI water, to perform the titrations. The results of this are shown in the data table below. Assuming 1:1 stoichiometry, what was the average concentration of NaOH determined by the student?

Trial # Mass of Acid (g) Volume of NaOH (mL)

1 0.7981 38.65

2 0.8044 39.10

3 0.8379 40.45

Please show work it helps so much

Answer 1

This is easy, because you already know that the stechiometry of reaction is 1:1. This means tha1t you can use the following expression:

M_aV_a = M_bV_b

We don't know the concentration and volume of the acid, but we do know the mass and MW. With this, we can calculate the moles (moles = MxV), so replacing this, we have:

moles a = M_bV_b   ----> M_b = moles a / V_b

so, first, calculate the moles of the acid for every mass:

moles 1 = 0.7981 / 204.22 = 3.91x10^-3 moles

moles 2 = 0.8044 / 204.22 = 3.94x10^-3 moles

moles 3 = 0.8379 / 204.22 = 4.1x10^-3 moles

Now with the above expression, we can calculate the concentration of the base:

M_b1 = 3.91x10^-3 / 0.03865 = 0.1012 M

M_b2 = 3.94x10^-3 / 0.03910 = 0.1008 M

M_b3 = 4.10x10^-3 / 0.04045 = 0.1014 M

The average would be:

M_b = 0.1012 + 0.1008 + 0.1014 / 3 = 0.1011 M

Hope this helps