Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of OH− ions in the solution after the reaction has gone to completion. Assume that there is no volume change when adding the grams of NaOH

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH =

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

of NaOH

Moles of NaOH left unreacted  in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

of hydroxide ion

The molarity of hydroxide ion in solution ;

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.