A 28.8 mL sample of HCl is titrated with NaOH. If 39.5 mL of 0.639 M NaOH is needed to reach the endpoint, what is the concentration (M) of the HCl solution

0.8764 M

Explanation:

Equation of the reaction:

NaOH + HCl --> NaCl + H2O

Number of moles = molar concentration × volume

= 0.039 × 0.639

= 0.0252 mol

By stoichiometry, 1 mole of HCl reacted with 1 mole of NaOH. Therefore, number of moles of HCl = 0.0252 mol

Molar concentration of HCl = number of moles ÷ volume

= 0.0252 ÷ 0.0288

= 0.8764 M of HCl.

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

i'm pretty sure its b i might be wrong though!