Titration problems

Question

A student performs a titration to determine the concentration of a solution of nitric acid, HNO3. The student starts with 45.0 mL of the acid in a flask and titrates it against a standard solution of sodium hydroxide, NaOH, with a known concentration of 1.47 M. The titration curve generated during the experiment is shown below.

1.Identify two ways the chemist knows this is a strong acid-strong base titration from the titration curve.

2.What is the concentration of the nitric acid solution?

Answer 1

Hello, Rochester,

The chemist knows this is a strong acid-base titration because 1) he/she read the safety sheet for each reagent, or 2) the titration "curve" is very steep, with only a small amount of the NaOH taking the pH from around 1 to 12 very quickly. A titration curve with a weak acid or base would show a more prolonged rise.

The concentration of the HNO₃ can be determined in two ways:

1. The Long Way, and
2. A Shortcut Calculation

a. The Long Way

The balanced equation for this titration is: HNO₃ + NaOH = NaNO₃ + H₂O

This tells us we need 1 mole of NaOH to neutralize each 1 mole of HNO₃. When the moles are equal, all of the OH^- groups are matched by H⁺ ions, so that [OH^-] = [H⁺]. This results in a neutral pH of 7. We can see from the titration curve that pH 7 is reached when 25ml of 1.47M NaOH is added.

Molarity is defined moles/liter. 25ml is 0.025 liters and 1.47M is the same as 1.47 moles/liter. So the number of moles NaOH required to reach pH 7 is:

(1.47moles/liter)*(0.025 liters) = 0.0368 moles of NaOH. That means we had 0.0368 moles of HNO₃ in the 45 ml of acid in the flask. 45 ml is 0.045 liter. Convert that into concentration (M) by dividing the moles HNO₃ by the volume of the solution, in liters:

(0.0368 moles acid)/(0.045 liters acid) = 0.8167M HNO₃ Round to 3 sig figs: 0.817M HNO₃.

b. The Shortcut Calculation

Since this is a strong acid-base titration, we can use the relationship M₁V₁ = M₂V₂ , where M and V are the Molarity and Volumes of the base (1) and acid (2). Both M1 and V1 are given to us (1.47M and 25 ml) and we know V2 (45 ml). Put thiose into the equation and solve for M2:

(1.47M)*(25.0 ml) = M2*(45.0 ml)

M2 = 0.816667M HNO₃, rounded to 0.817M for 3 sig figs, assuming I can actually read the titration curve to 3 places (25.0ml). But use 2 sig figs if you want to be safe. [Note that you can use either ml or liters in this calculation. The volume units cancel

I hope this helps,

Bob

Answer 2

1). You can tell it is a strong acid and strong base because of how steep the rise in the curve is, and also because at the equivalence point the pH is 7 (or very close to it). If either the acid or the base were weak, the slope would be more shallow and the pH would not be 7.

2). The reaction is HNO₃ + NaOH ==> NaNO₃ + H₂O

moles of NaOH = moles of HNO₃

moles NaOH = 1.47 mol/L x 0.025 L = 0.03675 moles NaOH (the 0.025 L is from the graph 25 ml = 0.025 L)

moles HCl = 0.03675 moles

Concentration of HCl = 0.03675 moles/0.045 L = 0.817 M (3 sig. figs.)