Question

A solution of an unknown acid is prepared by dissolving 5.968 g of the solid acid in sufficient DI water to make 350.00 mL of solution. 20.12 mL of a 0.1666 M NaOH solution are required to neutralize 10.00 mL of this acid solution?

Answer 1

Are you sure, that's your question?. Because in order to know if that volume is required to neutralize the solution, I need the molecular weight of the acid, and I need to know if the acid is monoprotic or diprotic.

In this case, I will assume the acid is monoprotic, and that you want to calculate the concentration of the acid along with the molecular weight.

In this case, we use the following equation: M_aV_a = M_bV_b and we solve for M_a:
M_a = 0.1666 M * (20.12/10) = 0.3352 M

Now, let's find out how many moles we have in the 350 mL of solution:
moles = 0.3352 * 0.350 = 0.1173 moles

Finally the molecular weight will be:
MW = 5.968 g / 0.1173 moles = 50.87 g/mol

Hope this helps