A chemist titrates 110.0mL of a 0.0516M nitric acid HNO3 solution with 0.3578M NaOH solution at 25°C . Calculate the pH at equivalence. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added.

pH = 5.1

Explanation:

To do this, let's write the general reaction of this:

HNO₃ + NaOH > NaNO₃ + H₂O

This equation is already balanced, and we can see that we have a 1:1 mole ratio between the acid and the base.

Now, the pH at the equivalence point would be the pH when this reaction has already reached the products, in other words, when this reaction has produced the products of the reaction.

As both reactants are strong reactants, we can assume that the moles of them are completely consumed in the reaction. The first thing we need to know is the moles of each reactant, in this way, we can know how much of each reactant was consumed and how much it was left.

To do this, let's calculate first the volume of the base used with the following expression:

M₁V₁ = M₂V₂

From this expression, let's calculate the volume of the base:

V₂ = M₁V₁/M₂

V₂ = 110 * 0.0516 / 0.3578 = 15.86 mL

We have the volume of the base, and the volume of base used, so the total volume would be: 110 + 15.86 = 125.86 mL

Now that we have this, let's calculate the moles of each reactant:

moles HNO₃ = 0.0516 * 0.110 = 0.005676 moles

moles NaOH = 0.3578 * 0.01586 = 0.005675 moles

Clearly we can see that we have more moles of acid than the moles of base, so if we do an ICE chart for this reaction we will have:

          HNO₃ + NaOH > NaNO₃ + H₂O

i)        0.005676   0.005675    0            0

e)       0.000001    0               0.005675  0.005675

This means that the limiting reactant is the base, that's why we have the reaction of this we the moles reacted of the base.

The remaining moles of the acid, will determine the pH of the solution at the equivalence point so, let's do the concentration of it:

[HNO₃] = 0.000001 / 0.12586 = 0.0000079 M

Finally the pH:

pH = -log[H⁺]

pH = -log(0.0000079)

pH = 5.1

And this would be the pH at the equivalence point

2b2t

explanation:

2b2t