5) D, For this question, write out the titration equation first: HNO3 + NaOH ---> H2O + NaNO3.
This is important because we want to see the stoichiometry between the acid and base. Now since this is a titration to completion, we know that all the H+ must be canceled out or neutralized by OH- to form water. This means we just need to find the moles of OH- because it is equal to the moles of H+.
We know that NaOH has a concentration of 0.1570M and a volume of 31.77mL and so knowing that molarity equals moles divided by volume in liters, we can solve for the moles of NaOH (and thus OH- since there is only one OH- in every molecule of NaOH). 0.1570M = x moles/0.03177L. Keep in mind that the volume must be in liters and I already converted the mL to L when I plugged the value in. Solve for x and you should get about 4.988x10^-3 moles of NaOH. Because every molecule of NaOH has 1 OH-, then we know that there must also be 4.988x10^-3 moles of OH-.
To titrate 4.988x10^-3 moles of OH-, we then need 4.988x10^-3 moles of H+. Each H+ can only be given off by every HNO3 and so therefore, we must have 4.988x10^-3 moles of HNO3 in the solution. Now, we are given the volume of the nitric acid so to find the molarity, divide the moles of HNO3 by the given 25.00mL (remember to change it to L).
Molarity = 4.988x10^-3 moles of HNO3/0.025L = about 0.1996M and that is your answer!
I hope this helped and always feel free to ask more questions! :)