Can you please help me with this problem?

Question

The percentage by weight of nitric acid HNO3 in a sample of concentrated nitric acid is to be determined.

A. Initially a NaOH solution was standardized by titration with a sample of potassium hydrogen phthalate, KHC8H4O4, a monoprotic acid often used as a primary standard. A sample of pure KHC8H4O4 weighing 1.518 grams was dissolved in water and tritrated with the NaOH solution. To reach the equivalence point, 26.90mL of base was required. Calculate the molarity of the NaOH

Answer 1

It is important to understand what is occurring during a titration at the equivalence point. In this problem, the monoprotic weak acid (KHC₈H₄O₄) is being titrated with a strong base (NaOH). At the equivalence point for titrations, the amount of moles of H⁺ is equal to the amount of moles of OH^- in solution. Since we know we have 1.518 grams of weak acid, KHC₈H₄O₄, we can convert to moles of weak acid (by using the periodic table to find the molar mass of the acid) which will also be the amount of moles of base in the solution. 

 

moles of acid = moles of base at the equivalence point

 

Molarity of a solution has units of moles per liter. We know that there is 26.90 mL of base and we can calculate the amount of moles of base. 

 

molar mass of weak acid= (39.10 amu K) + 5(1.008 amu H) + 8(12.01 amu C) + 4(16.00 amu O) = 204.22 grams per mole of weak acid. 

 

find moles of weak acid= 1.518 grams x 1 mole/ 204.22 grams = 7.433 x 10^-3 moles of weak acid 

 

7.433 x 10^-3 moles of acid = 7.433 x 10^-3 moles of base

 

find the molarity of base= 7.433 x 10^-3 moles / 0.02690 L base = 0.2763 M NaOH

 

Answer 2

Please see my previous answer:

 

[NaOH] = 0.007433 moles/0.02690 L = 0.2763 M