What volume of a 0.311 M hydrochloric acid solution is required to neutralize 13.9 mL of a 0.110 M potassium hydroxide solution?

4.9 mL

Explanation:

To answer this question we will use the following equation:

C₁V₁=C₂V₂

Where the subscript 1 refers to hydrochloric acid, while 2 refers to potassium hydroxide. Meaning:

C₁ = 0.311 MV₁ = ?C₂ = 0.110 MV₂ = 13.9 mL

So we solve for V₁:

0.311 M * V₁= 0.110 M * 13.9 mL V₁ = 4.9 mL

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.