You are performing an acid-base neutralization reaction in the laboratory to determine the concentration of an unknown base. You are supposed to titrate it with a monoprotic acid, but your lab partner accidentally fills your buret with sulfuric acid, a diprotic acid, with the same concentration as the acid called for in the experiment. How will the volume of diprotic acid compare to the volume of monoprotic acid you would have used

Answer:

Volume is reduced to half

Explanation:

Acid base titration are commonly used reactions in a lab, and are ofter used to get pH or different kind of solutions.

The neutralization of an acid base reaction is reached, when the solution (having added an indicator previously) changes its original color. chemically speaking, this occurs when the number of moles of the acid and the base are balanced and equal. In other words the following:

n₁ = n₂   (1)

This expression can also be expressed in function of concentration and volume:

M₁V₁ = M₂V₂   (2)

From here, solving for V₁:

V₁ = M₂V₂  / M₁

Now, this expression is true only when we have the same kind of substance that can lose or gain the same number of hydrogens.

Lets suppose that we have as base NaOH (Monoprotic base) and HCl (monoprotic acid), the titration reaction would be:

NaOH + HCl --------> NaCl + H₂O

As both of the species are monoprotic, the number of moles are the same  when they reach the equilibrium, so, expression  (2) can be used, and calculate volume or concentrations.

However, in this case, a partner made a mistake and use a diprotic acid, in this case, H₂SO₄, In this case, things chance because H₂SO₄ is diprotic, meaning that we need to dissociate two hydrogens in equilibrium, therefore, expression (2) would be something like this.

Acid: 1; Base: 2

H₂SO₄ + 2NaOH ------> Na₂SO₄ + H₂O

nH₂SO₄ = n₁ = 1

nNaOH = n₂ = 2

n₁/n₂ = 1/2

2n₁ = n₂   (3)

Writting this, in function of concentration and volume, it would be:

2M₁V₁ = M₂V₂   (4)

From here, if we solve for the volume of the acid (V₁):

V₁ = M₂V₂ / 2M₁

Therefore, according to this expression, we can see that the volume required of the acid would be half the volume required of the monoprotic acid. For example, if we need 50 mL of Chloridic acid to reach the equivalence point with NaOH, then, with H₂SO₄ it will only need 25 mL. This, of course, assuming that concentrations are the same, and volume of the base used, the same.

Hope this helps

Consider the following data on some weak acids and weak bases
acid
Ka
name formula
acetic acid
HCH3CO2
1.8 x10−5
hydrocyanic acid
HCN
4.9 x 10−10
base
Kb
name formula
pyridine
C5H5N
1.7 x 10−9
ammonia
NH3
1.8 x 10−5
Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on.
a. 0.1M NaCH3CO2
b. 0.1M NH4Br
c. 0.1M NaBr
d. 0.1M KCN

Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expression for this reaction. Then create an ICE chart. Since we know the equilibrium concentration of the silver ion, we can solve for Ksp.Does it agree with the literature value