Tag: acids, bases and salts

Questions Related to acids, bases and salts

${ H } _{ 3 }{ PO } _{ 4 }$ is a tribasic acid and one of its salt is $Na{ H } _{ 2 }{ PO } _{ 4 }$. What volume of 1M NaOH solution should be added to 12g $Na{ H } _{ 2 }{ PO } _{ 4 }$ to convert it into $Na _{ 3 }{ PO } _{ 4 }$? (at.wt of P=31)

  1. 100 mL

  2. 200 mL

  3. 80 mL

  4. 300 mL


Correct Option: B
Explanation:

$NaH _2PO _4$ + $ 2 NaOH $ $ \rightarrow Na _3PO _4 + H _2O $

From the above reaction we can say that,
1 mol of $NaH _2PO _4$ reacts with 2 mol of NaOH.
Molar mass of $NaH _2PO _4$  = 120 g / mol
12g of $NaH _2PO _4$ contains = 0.1 mol
So the number of moles of NaOH required = 0.2 mol
Molarity of NaOH = 1M
1 mol = 1000 ml
0.2 mol = ?
So, the volume of 1M NaOH solution = 200 ml

Which of the following has the highest melting point ?

  1. $BeCl _2$

  2. $LiCl$

  3. $AlCl _3$

  4. $NaCl$


Correct Option: D
Explanation:

Sodium and magnesium chlorides are solids with high melting and boiling points because of the large amount of heat which is needed to break the strong ionic attractions.
$NaCl$ having the highest melting point as the ionic bond is too strong between $Na$ and $Cl$.

Equal volumes of $1\ M$ $HCl$ and $1\ M$ $H _2SO _4$ are neutralized by dilute NaOH solution liberating $x$ kcal and $y$ kcal heat. Which of the following is correct?

  1. $x=y$

  2. $x=0.5y$

  3. $x=0.4y$

  4. None of these


Correct Option: B
Explanation:
Reaction involved are:
$HCl+NaOH \rightarrow NaCl+H _2O, \Delta H _1=xkcal$
$H _2SO _4+2NaOH \rightarrow Na _2SO _4+2H _2O, \Delta H _1=ykcal$
(i) acid-base neutralization reaction for HCl is represented as:
$H^++OH^- \rightarrow H _2O, \Delta H _1=xkcal$
(ii) acid-base neutralization reaction for $H _2SO _4$ is represented as:
$2H^++2OH^- \rightarrow 2H _2O, \Delta H _1=ykcal$
since reaction (ii) is twice of reaction (i), so is the enthalpies
thus $2x=y$
or $x=0.5y$

Which of the following salts is most basic in aqueous solution?

  1. ${\text{C}}{{\text{H}} _{\text{3}}}{\text{COOK}}\,$

  2. ${\text{FeC}}{{\text{l}} _{\text{3}}}\,\,$

  3. ${\text{Pb(C}}{{\text{H}} _{\text{3}}}{\text{COOK}}{{\text{)}} _{\text{2}}}\,$

  4. ${\text{Al}}{\left( {{\text{CN}}} \right) _{\text{3}}}$


Correct Option: A

A and B are two salts. A with dil. HCI and A and B with conc. ${H _2}S{O _4}$react to give reddish brown vapours, hence A and B respectively are:

  1. $NaN{O _3},NaBr$

  2. $NaBr,NaN{O _3}$

  3. $NaBr,NaN{O _2}$

  4. $NaN{O _{2,}}NaBr$


Correct Option: B

A and B are two salts. A with dil. HCI and A and B with conc. ${H _2}S{O _4}$  react  to give reddish brown  vapour, hence A and B respectively are.

  1. $NaN{O _3},NaBr$

  2. $NaBr,NaN{O _3}$

  3. $NaBr,NaN{O _2}$

  4. $NaN{O _{2,}}NaBr$


Correct Option: B

In the reaction, $BaCl _{2}(aq)+K _{2}SO _{4}(aq)\rightarrow BaSO _4(s)+2KCl(aq)$, which is not a spectator ion?

  1. $K^{+}$

  2. $Cl^{-}$

  3. $Ba^{2+}$ and $SO _{4}^{-2}$

  4. Both $A$ and $B$


Correct Option: C
Explanation:

Ions that are not involved in the formation of insoluble solid are called as spectator ions.


Therefore, $Ba^{2+}$ and $SO _{4}^{2-}$ ions form insoluble solid.

Hence they are not spectator ions.


Option C is correct.

We can obtain zinc sulphate from an active metal.

  1. True

  2. False


Correct Option: A
Explanation:

$Zn+{ H } _{ 2 }{ SO } _{ 4 }(dil.)\rightarrow { ZnSO } _{ 4 }+{ H } _{ 2 }$

Which of the following methods is generally used for preparing $Zn(NO _3) _2$ salt?


1 : Neutralisation - insoluble base & dilute acid  
2 : Neutralisation - alkali & dilute acid      

3 : Simple displacement - active metal & dilute acid  
4 : Direct combination                     

5 : Precipitation [double decomposition]

  1. 2

  2. 3

  3. 4

  4. 5


Correct Option: B
Explanation:

Zinc metal with dilute nitric acid gives zinc nitrate.

So, simple displacement-active metal & dilute acid is the correct option for this.

$Zn+2HNO _3 \rightarrow Zn(NO _3) _2+H _2$

Hence, the correct option is $(B)$.

State which of the following methods is generally used for preparing the $FeS$ salts
1: Neutralisation-insoluble base & dilute acid.
2: Neutralisation-alkali & dilute acid.
3: Simple displacement-active metal & dilute acid.
4: Direct combination.
5: Precipitation [double decomposition].
  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: D
Explanation:


Iron(II) sulfide is obtained by heating of iron and sulfur which is a direct combination.
$Fe+S \longrightarrow FeS$


Hence, the correct option is $(D)$