Question

question 5 (start on a new page)
sulphuric acid is a strong acid and reacts with water in two steps as shown in the reactions below
h2so4 + h2o → hso4 -+ h3o+
hso4 -+ h2o → so4 2-+ h3o+
5.1 what is meant by the term diprotic acid?
5.2 identify the ampholyte in the above equations.
5.3 write down any one of the conjugate acid - base pairs in the first reaction. clearly identify the acid and the base.
5.4 give an example of any salt that will form with the acid in the second reaction.
5.5 what colour will bromothymol blue be if it is added to the second reaction?
5.6 a solution of potassium hydroxide (koh) is prepared by dissolving 3.36g crystals of koh in 250 cm3 of solution.
5.6.1 is koh a weak or strong base? explain your answer.
5.6.2 calculate the concentration of the potassium hydroxide solution.
5.7 25 cm3 of potassium hydroxide solution of concentration 0.25 mol.dm - 3 completely neutralises a dilute solution of sulphuric acid (h2so4) in a flask. the incomplete equation below represents the reaction that takes place:
2koh (aq)+h2so4 (aq)→k2so4(aq)+d
5.7.1 write down the name of the salt formed.
5.7.2 write down the formula of compound d.
5.7.3 calculate the mass of sulphuric acid in the flask.

Explanation:

Step1: Define diprotic acid

A diprotic acid is an acid that can donate two protons ($H^+$) per molecule in successive ionization steps.

Step2: Identify ampholyte

An ampholyte can act as both an acid and a base. In the given equations, $HSO_4^-$ can donate a proton (act as an acid) and also accept a proton (act as a base), so the ampholyte is $HSO_4^-$.

Step3: Identify conjugate - acid base pair

In the first reaction $H_2SO_4 + H_2O
ightarrow HSO_4^-+H_3O^+$, the conjugate - acid base pair is $H_2SO_4$ (acid) and $HSO_4^-$ (base).

Step4: Provide salt example

In the second reaction $HSO_4^- + H_2O
ightarrow SO_4^{2 -}+H_3O^+$, an example of a salt that can form is $Na_2SO_4$ (sodium sulfate).

Step5: Determine bromothymol blue color

The second reaction produces $H_3O^+$, making the solution acidic. Bromothymol blue is yellow in acidic solutions, so the color is yellow.

Step6.1: Classify KOH as base

$KOH$ is a strong base because it completely dissociates in water to produce $K^+$ and $OH^-$ ions: $KOH(s)
ightarrow K^+(aq)+OH^-(aq)$.

Step6.2: Calculate KOH concentration

First, find the number of moles of $KOH$:
The molar mass of $KOH$ is $M = 39 + 16+ 1=56\ g/mol$.
The number of moles $n=\frac{m}{M}=\frac{3.36\ g}{56\ g/mol}=0.06\ mol$.
The volume $V = 250\ cm^3=0.25\ dm^3$.
The concentration $c=\frac{n}{V}=\frac{0.06\ mol}{0.25\ dm^3}=0.24\ mol/dm^3$.

Step7.1: Name the salt

From the reaction $2KOH(aq)+H_2SO_4(aq)
ightarrow K_2SO_4(aq)+D$, the name of the salt is potassium sulfate.

Step7.2: Determine compound D

Based on the law of conservation of mass and the reaction of an acid and a base to form a salt and water, compound D is $H_2O$.

Step7.3: Calculate mass of sulfuric acid

The balanced equation is $2KOH(aq)+H_2SO_4(aq)
ightarrow K_2SO_4(aq)+2H_2O$.
The number of moles of $KOH$:
$V = 25\ cm^3 = 0.025\ dm^3$, $c = 0.25\ mol/dm^3$, so $n_{KOH}=cV=0.25\ mol/dm^3\times0.025\ dm^3 = 0.00625\ mol$.
From the mole - ratio in the balanced equation, the mole - ratio of $KOH$ to $H_2SO_4$ is $2:1$. So the number of moles of $H_2SO_4$, $n_{H_2SO_4}=\frac{0.00625\ mol}{2}=0.003125\ mol$.
The molar mass of $H_2SO_4$ is $M = 2\times1+32 + 4\times16=98\ g/mol$.
The mass of $H_2SO_4$ is $m = n\times M=0.003125\ mol\times98\ g/mol = 0.30625\ g$.

Answer:

5.1: An acid that can donate two protons per molecule in successive ionization steps.
5.2: $HSO_4^-$
5.3: Acid: $H_2SO_4$, Base: $HSO_4^-$
5.4: $Na_2SO_4$
5.5: Yellow
5.6.1: Strong base; it completely dissociates in water to produce $K^+$ and $OH^-$ ions.
5.6.2: $0.24\ mol/dm^3$
5.7.1: Potassium sulfate
5.7.2: $H_2O$
5.7.3: $0.30625\ g$