A27.86 ml sample of 0.1744 m hno3 is tirades with 29.4ml of a job solution. what is the molarity of the koh

0.165 mol·L⁻¹

Explanation:

1. Write the chemical equation for the reaction.

HNO₃ + KOH ⟶ KNO₃ + H₂O

2. Calculate the moles of HNO₃

c = n/V                                               Multiply each side by V and transpose

n = Vc

V = 0.027 86 L

c = 0.1744 mol·L⁻¹                             Calculate the moles of HNO₃

Moles of HNO₃ = 0.027 86 × 0.1744

Moles of HNO₃ = 4.859 × 10⁻³ mol HNO₃

3. Calculate the moles of KOH

1 mol KOH ≡ 1 mol HNO₃                 Calculate the moles of KOH

Moles of KOH = 4.859 × 10⁻³× 1/1

Moles of KOH = 4.859 × 10⁻³ mol KOH

4. Calculate the molar concentration of the KOH

V = 29.4 mL = 0.0294 L                   Calculate the concentration

c = 4.859 × 10⁻³/0.0294

c = 0.165 mol·L⁻¹

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.