Acids and Bases Questions in English

(C) The concept of conjugate acid-base pairs was proposed by the $Lowry-Bronsted$ theory.
According to this theory,an acid is a proton $(H^+)$ donor,and a base is a proton $(H^+)$ acceptor.
To find the conjugate base of a species,we remove one proton $(H^+)$ from it.
For the given species $H_2PO_4^-$,the removal of one proton $(H^+)$ is represented as:
$H_2PO_4^- \rightarrow HPO_4^{2-} + H^+$
Thus,the conjugate base of $H_2PO_4^-$ is $HPO_4^{2-}$.

(A) An acid salt is formed by the partial replacement of replaceable hydrogen atoms from a polyprotic acid by a metal or a positive ion.
$H_3PO_2$ (hypophosphorous acid) is a monobasic acid because it contains only one $P-OH$ bond.
Since it has only one replaceable hydrogen atom,it cannot form an acid salt.
$NaH_2PO_2$ is a normal salt of $H_3PO_2$ and $NaOH$.
Therefore,$NaH_2PO_2$ is not an acid salt.

(A) The relative acidic strength depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
The order of stability of conjugate bases is $CH_3COO^- > CN^- > OH^- > C_2H_5O^-$.
Consequently,the order of acidic strength is $CH_3COOH > HCN > HOH > C_2H_5OH$.
Thus,the correct option is $A$.

(D) conjugate acid is formed when a base accepts a proton $(H^+)$.
Compound $(A)$ is $4$-aminocyclohexan-$1$-ol,which contains two basic sites: the amino group $(-NH_2)$ and the hydroxyl group $(-OH)$.
$1$. The nitrogen atom in the $-NH_2$ group has a lone pair of electrons,making it a Lewis base. It can accept a proton to form the ammonium ion $(-NH_3^+)$.
$2$. The oxygen atom in the $-OH$ group also has lone pairs of electrons and can be protonated in the presence of a strong acid to form the oxonium ion $(-OH_2^+)$.
Therefore,both the protonated nitrogen species and the protonated oxygen species are conjugate acids of $(A)$.

(B) In the given reaction $CH_3COOH + HCl \rightleftharpoons Cl^{\Theta} + CH_3COOH_2^{\oplus}$,$HCl$ acts as a strong acid because it donates a proton $(H^{\oplus})$ to $CH_3COOH$.
When an acid loses a proton,the remaining species is its conjugate base.
Therefore,the conjugate base of $HCl$ is $Cl^{\Theta}$.

(B) The conjugate acid of a base is formed by the addition of a proton $(H^{+})$ to the base.
For the base $HPO_4^{2-}$,the conjugate acid is obtained by adding one $H^{+}$ ion:
$HPO_4^{2-} + H^{+} \rightarrow H_2PO_4^{-}$
Therefore,the conjugate acid of $HPO_4^{2-}$ is $H_2PO_4^{-}$.

(A) The strength of a conjugate base is inversely proportional to the strength of its conjugate acid.
The conjugate acids of the given anions are $HCN$,$HCl$,$HI$,and $HBr$.
Among these,$HI$ is the strongest acid and $HCN$ is the weakest acid.
Since $HCN$ is the weakest acid,its conjugate base $CN^{-}$ is the strongest Bronsted base.

(C) $Bronsted$ base is defined as a substance that can accept a proton $(H^+)$.
$HCO_3^-$ can accept a proton to form $H_2CO_3$.
$PH_3$ has a lone pair on the $P$ atom and can accept a proton.
$H_2O$ can accept a proton to form $H_3O^+$.
$BF_3$ is an electron-deficient molecule with an incomplete octet (only $6$ electrons around $B$). It acts as a $Lewis$ acid by accepting an electron pair,not as a $Bronsted$ base.

(C) Lewis base is defined as a substance that can donate a lone pair of electrons.
$H_2O$ has two lone pairs on the oxygen atom,$NH_3$ has one lone pair on the nitrogen atom,and $F^{-}$ has four lone pairs on the fluorine atom.
$CH_4$ (methane) has a stable octet configuration with all valence electrons involved in bonding,and it has no lone pairs to donate.
Therefore,$CH_4$ cannot act as a Lewis base.

(D) According to the $Lewis$ acid-base theory:
$1$. $A$ $Lewis$ acid is defined as an electron pair acceptor.
$2$. $A$ $Lewis$ base is defined as an electron pair donor.
Therefore,the correct definition for a $Lewis$ acid is a lone pair acceptor.

(D) An amphoteric species is one that can act as both an acid (proton donor) and a base (proton acceptor).
For an anion to be amphoteric,it must contain at least one hydrogen atom that can be donated as a proton $(H^+)$ and have a negative charge to accept a proton.
In the case of $HCO_3^-$,it can donate a proton to form $CO_3^{2-}$ $(HCO_3^- \rightarrow H^+ + CO_3^{2-})$ and it can accept a proton to form $H_2CO_3$ $(HCO_3^- + H^+ \rightarrow H_2CO_3)$.
Therefore,$HCO_3^-$ is an amphoteric anion.

(C) An amphoteric species is one that can act as both a Bronsted-Lowry acid (donate a proton) and a Bronsted-Lowry base (accept a proton).
$(I) \, H_2O$: Can accept a proton to form $H_3O^+$ and donate a proton to form $OH^-$. Thus,it is amphoteric.
$(II) \, NH_2^-$: Can only accept a proton to form $NH_3$. It cannot donate a proton. Thus,it is not amphoteric.
$(III) \, H_2PO_4^-$: Can accept a proton to form $H_3PO_4$ and donate a proton to form $HPO_4^{2-}$. Thus,it is amphoteric.
$(IV) \, HCO_3^-$: Can accept a proton to form $H_2CO_3$ and donate a proton to form $CO_3^{2-}$. Thus,it is amphoteric.
Therefore,$(I), \, (III),$ and $(IV)$ are amphoteric species.

(B) conjugate base is formed by the removal of a proton $(H^+)$ from an acid.
For the complex $[Al(H_2O)_3(OH)_3]$,the removal of one $H^+$ ion from one of the water ligands results in the formation of an $OH^-$ group.
Reaction: $[Al(H_2O)_3(OH)_3] \rightarrow [Al(H_2O)_2(OH)_4]^- + H^+$.
Therefore,the conjugate base is $[Al(H_2O)_2(OH)_4]^-$,which corresponds to option $B$.

(C) $BF_3$ (Boron trifluoride) has an incomplete octet around the central Boron atom,making it an electron-deficient species.
According to the Lewis acid-base theory,an acid is an electron pair acceptor.
Since $BF_3$ can accept a lone pair of electrons to complete its octet,it acts as a Lewis acid.

(D) The strength of a Bronsted base is inversely proportional to the strength of its conjugate acid.
$1$. The conjugate acids of the given bases are: $H^{-} \rightarrow H_2$,$CH_3^{-} \rightarrow CH_4$,$CH_3O^{-} \rightarrow CH_3OH$,and $Cl^{-} \rightarrow HCl$.
$2$. The acid strength order is: $HCl > CH_3OH > H_2O > CH_4 > H_2$.
$3$. Since $HCl$ is the strongest acid among these,its conjugate base,$Cl^{-}$,must be the weakest Bronsted base.
Therefore,the correct option is $D$.

(A) According to the Bronsted-Lowry theory,an acid is a proton donor and a base is a proton acceptor.
When ammonia $(NH_3)$ accepts a proton $(H^+)$,it forms the ammonium ion $(NH_4^+)$.
Since $NH_4^+$ can donate a proton to form $NH_3$,it acts as a conjugate acid of the base $NH_3$.

(C) The proton-accepting capacity is determined by the basicity of the molecule,which depends on the availability of the lone pair of electrons on the central atom.
$NH_3$ has a nitrogen atom with a lone pair of electrons. Nitrogen is more electronegative than phosphorus,but the lone pair in $NH_3$ is more available for donation compared to $PH_3$ because the $N-H$ bond is shorter and the lone pair is in a smaller orbital.
$H_2O$ and $H_2S$ are less basic than $NH_3$ and $PH_3$ because oxygen and sulfur are more electronegative,holding their lone pairs more tightly.
Therefore,$NH_3$ is the strongest base among the given options and has the highest proton-accepting capacity.

(A) $Br\text{ø}nsted$ acid is a proton $(H^+)$ donor,and a $Br\text{ø}nsted$ base is a proton $(H^+)$ acceptor.
Species that can act as both are called amphoteric.
For $HSO_4^-$,it can donate a proton to form $SO_4^{2-}$ $(HSO_4^- \rightarrow H^+ + SO_4^{2-})$ and accept a proton to form $H_2SO_4$ $(HSO_4^- + H^+ \rightarrow H_2SO_4)$.
Therefore,$HSO_4^-$ is amphoteric.

(C) For a diprotic acid $H_2A$,the ionization occurs in two steps:
$1$. $H_2A(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HA^-(aq)$ with constant $Ka_1$.
$2$. $HA^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^{2-}(aq)$ with constant $Ka_2$.
In the second step,the removal of a proton $(H^+)$ from a negatively charged species $(HA^-)$ is much more difficult due to electrostatic attraction between the positive proton and the negative ion. Therefore,$Ka_1$ is always greater than $Ka_2$ ($Ka_1 > Ka_2$ or $Ka_2 < Ka_1$).

(B) The dissociation of a weak acid like $HCN$ requires energy to break the $H-CN$ bond. Therefore,the process is endothermic.

(B) The enthalpy of neutralization for a strong acid and a strong base is the heat released when $1$ mole of $H^+$ ions from the acid reacts with $1$ mole of $OH^-$ ions from the base to form $1$ mole of water. This value is constant for all strong acid-strong base reactions and is equal to $-57.3 \ kJ/mol$.

(D) The enthalpy of neutralization of a weak acid with a strong base is given by $\Delta H_{neut} = -57.3 + \Delta H_{ion}$,where $\Delta H_{ion}$ is the enthalpy of ionization of the weak acid.
Since the enthalpy of neutralization is less exothermic for $HCN$ $(12.13 \ kJ \ equi^{-1})$ compared to $H_2S$ $(15.9 \ kJ \ equi^{-1})$,it implies that more energy is required to ionize $HCN$ than $H_2S$.
This indicates that $HCN$ is a weaker acid than $H_2S$.
Since $pK_a = -\log(K_a)$,a weaker acid has a smaller $K_a$ value and consequently a larger $pK_a$ value.
Therefore,the $pK_a$ of $HCN$ is greater than the $pK_a$ of $H_2S$.

(D) The rise in temperature of the solution during neutralization is directly proportional to the enthalpy of neutralization.
For strong acid and strong base,the enthalpy of neutralization is $-57.1 \ kJ \ mol^{-1}$.
For weak acids,some energy is consumed in the dissociation of the weak acid,resulting in a lower enthalpy of neutralization compared to strong acids.
However,the question asks for the maximum rise in temperature. Among the given options,$HNO_3$ is a strong acid,while $HCN$,$CH_3COOH$,and $H_2S$ are weak acids.
Since the enthalpy of neutralization for a strong acid $(HNO_3)$ is the highest,it will release the maximum amount of heat,leading to the maximum rise in temperature.

(B) The heat of neutralization for the reaction between a strong acid and a strong base is defined as the enthalpy change when $1 \ mole$ of $H^+$ ions reacts with $1 \ mole$ of $OH^-$ ions to form $1 \ mole$ of water.
This value is constant for all strong acid-strong base neutralizations and is equal to $-57.3 \ kJ \ mol^{-1}$.
The reaction is represented as: $H^+_{(aq)} + OH^-_{(aq)} \to H_2O_{(l)}$,$\Delta H = -57.3 \ kJ \ mol^{-1}$.

(B) protonic acid is a substance that can donate a proton ($H^+$ ion) in an aqueous solution.
$SO_2(OH)_2$ is $H_2SO_4$,which is a strong protonic acid.
$PO(OH)_3$ is $H_3PO_4$,which is a protonic acid.
$SO(OH)_2$ is $H_2SO_3$,which is a protonic acid.
$B(OH)_3$ is boric acid $(H_3BO_3)$. It acts as a Lewis acid because it accepts a lone pair of electrons from $OH^-$ ions in water to form $[B(OH)_4]^-$,rather than donating a proton. Therefore,it is not a protonic acid.

(B) $Al^{3+}$ ions react with $NaOH$ to form $Al(OH)_3$ initially.
However,$Al(OH)_3$ is amphoteric in nature and dissolves in excess $NaOH$ to form a soluble complex,$[Al(OH)_4]^-$.
To avoid the formation of this soluble complex and ensure complete precipitation of $Al(OH)_3$,a weaker base like aqueous ammonia $(NH_4OH)$ is used.
$NH_4OH$ provides a lower concentration of $OH^-$ ions,which is sufficient to precipitate $Al(OH)_3$ but not enough to dissolve it.

(D) $Lewis$ acid is a substance that can accept a pair of non-bonding electrons.
$AlCl_3$,$SnCl_4$,and $FeCl_3$ are electron-deficient compounds (incomplete octet or vacant d-orbitals) and act as $Lewis$ acids.
$AlCl_3 \cdot 6H_2O$ is a hydrated salt where the $Al^{3+}$ ion is already coordinated with six water molecules,satisfying its coordination requirement. Therefore,it does not act as a $Lewis$ acid in this form.

(C) The heating of iron sulfide $(FeS_2)$ in air produces sulfur dioxide $(SO_2)$.
$4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2$
Sulfur dioxide reacts with water to form sulfurous acid $(H_2SO_3)$.
$SO_2 + H_2O \rightarrow H_2SO_3$
Sulfurous acid $(H_2SO_3)$ is a diprotic acid,meaning it has two ionizable hydrogen atoms.
Therefore,the basicity of $H_2SO_3$ is $2$.

(A) $1$. $A$ species is basic if it can accept a proton $(H^+)$.
$2$. $A$ species is a reducing agent if it can be oxidized to a higher oxidation state.
$3$. In $SO_3^{2-}$,the oxidation state of $S$ is $+4$. It can be oxidized to $SO_4^{2-}$ $(+6)$ and it can accept $H^+$ to form $HSO_3^-$.
$4$. $SO_4^{2-}$ is the highest oxidation state of $S$ $(+6)$,so it cannot act as a reducing agent.
$5$. $S_2O_4^{2-}$ (dithionite) is a strong reducing agent but is less commonly cited as a simple base compared to $SO_3^{2-}$.
$6$. $HSO_4^-$ is acidic in nature.
$7$. Therefore,$SO_3^{2-}$ is the correct answer.

(A) The $pK_a$ value is inversely proportional to the acid strength $(pK_a = -\log K_a)$.
Stronger acids have lower $pK_a$ values.
The order of acid strength is: $HClO_4$ (strongest) $> HNO_3 > H_2CO_3 > B(OH)_3$ (weakest).
Therefore,the increasing order of $pK_a$ values is: $HClO_4 < HNO_3 < H_2CO_3 < B(OH)_3$.

(A) The basicity of an anion is inversely proportional to the stability of its conjugate acid.
The conjugate acids of the given anions are $HF$,$HCl$,$HBr$,and $HI$.
Among these,$HI$ is the strongest acid because the $H-I$ bond is the weakest due to the large size of the $I^-$ ion.
Conversely,$HF$ is the weakest acid among the hydrohalic acids.
Since the strength of the conjugate acid is inversely related to the basicity of the conjugate base,the weakest acid $(HF)$ will have the strongest conjugate base $(F^-)$.
Therefore,$F^-$ is the most basic among the given options.

(D) For $HNO_3$ to act as a base,it must accept a proton $(H^+)$ from another substance.
This occurs in the presence of a stronger acid,such as $H_2SO_4$ or $HClO_4$.
Among the given options,$HIO_4$ (periodic acid) is the strongest acid because the iodine atom is in a high oxidation state $(+7)$ and is bonded to four highly electronegative oxygen atoms,which stabilizes the conjugate base $(IO_4^-)$ through resonance and inductive effects.
Therefore,$HIO_4$ is the strongest acid among the choices provided.

(C) When potassium oxide $(K_2O)$ is added to water,it undergoes a hydrolysis reaction to form potassium hydroxide $(KOH)$.
The chemical equation is: $K_2O(s) + H_2O(l) \rightarrow 2KOH(aq)$.
Potassium hydroxide is a strong base that dissociates completely in water to produce hydroxide ions $(OH^{-})$.
The presence of these $OH^{-}$ ions makes the solution basic.

(C) The conjugate base of a Bronsted acid is formed by the removal of a proton $(H^{+})$ from the acid.
For $H_{2}O$: $H_{2}O - H^{+} \rightarrow OH^{-}$.
For $HF$: $HF - H^{+} \rightarrow F^{-}$.
Therefore,the conjugate bases are $OH^{-}$ and $F^{-}$ respectively.

(C) Bronsted acid is a proton donor,and a Bronsted base is a proton acceptor.
Species that can act as both are called amphoteric.
$HCO_3^-$,$NH_3$,and $HSO_4^-$ possess both a proton to donate and a lone pair/negative charge to accept a proton.
$HCl$ can only donate a proton $(HCl \rightarrow H^+ + Cl^-)$ but cannot accept a proton to form $H_2Cl^+$,thus it cannot act as a Bronsted base.

(A) The conjugate base of a Bronsted acid is formed by removing one proton $(H^{+})$ from the acid molecule.
For $HF$: $HF - H^{+} \rightarrow F^{-}$.
For $H_{2}SO_{4}$: $H_{2}SO_{4} - H^{+} \rightarrow HSO_{4}^{-}$.
For $HCO_{3}^{-}$: $HCO_{3}^{-} - H^{+} \rightarrow CO_{3}^{2-}$.
Therefore,the corresponding conjugate bases are $F^{-}$,$HSO_{4}^{-}$,and $CO_{3}^{2-}$ respectively.

(A) The conjugate acid is formed by adding one proton $(H^{+})$ to the given Bronsted base.
For $NH_{2}^{-}$,adding $H^{+}$ gives $NH_{3}$.
For $NH_{3}$,adding $H^{+}$ gives $NH_{4}^{+}$.
For $HCOO^{-}$,adding $H^{+}$ gives $HCOOH$.
Therefore,the corresponding conjugate acids are $NH_{3}$,$NH_{4}^{+}$ and $HCOOH$ respectively.

$A$ Bronsted acid is a proton $(H^{+})$ donor,and a Bronsted base is a proton $(H^{+})$ acceptor. The conjugate acid is formed by adding a proton to the species,while the conjugate base is formed by removing a proton from the species. The results are summarized in the table below:

Species	Conjugate Acid	Conjugate Base
$H_{2}O$	$H_{3}O^{+}$	$OH^{-}$
$HCO_{3}^{-}$	$H_{2}CO_{3}$	$CO_{3}^{2-}$
$HSO_{4}^{-}$	$H_{2}SO_{4}$	$SO_{4}^{2-}$
$NH_{3}$	$NH_{4}^{+}$	$NH_{2}^{-}$

(N/A) $(a) HO^{-}$ is a Lewis base because it has lone pairs of electrons that it can donate.
$(b) F^{-}$ is a Lewis base because it has lone pairs of electrons that it can donate.
$(c) H^{+}$ is a Lewis acid because it has an empty orbital and can accept a lone pair of electrons.
$(d) BCl_{3}$ is a Lewis acid because the boron atom has an incomplete octet (only $6$ electrons in its valence shell) and can accept a lone pair of electrons.

Auto-protolysis (self-ionization) of water is a chemical reaction in which two water molecules react to produce a hydroxide ion $(OH^{-})$ and a hydronium ion $(H_{3}O^{+})$. The reaction involved can be represented as:
$H_{2}O_{(l)} + H_{2}O_{(l)} \leftrightarrow H_{3}O_{(aq)}^{+} + OH_{(aq)}^{-}$
Auto-protolysis of water indicates its amphoteric nature,i.e.,its ability to act as an acid as well as a base.
The acid-base reaction can be written as:
$(H_{2}O_{(l)})_{\text{acid}} + (H_{2}O_{(l)})_{\text{base}} \leftrightarrow (H_{3}O_{(aq)}^{+})_{\text{conjugate acid}} + (OH_{(aq)}^{-})_{\text{conjugate base}}$

(N/A) $H_{2}SO_{4(aq)} + H_{2}O_{(l)} \to H_{3}O_{(aq)}^{+} + HSO_{4(aq)}^{-}; \quad K_{a_{1}} > 10$
$HSO_{4(aq)}^{-} + H_{2}O_{(l)} \to H_{3}O_{(aq)}^{+} + SO_{4(aq)}^{2-}; \quad K_{a_{2}} = 1.2 \times 10^{-2}$
It can be observed that $K_{a_{1}} >> K_{a_{2}}$.
This is because a neutral $H_{2}SO_{4}$ molecule has a much higher tendency to lose a proton compared to the negatively charged $HSO_{4}^{-}$ ion. The electrostatic attraction between the positively charged proton and the negatively charged $HSO_{4}^{-}$ ion makes the removal of the second proton significantly more difficult.

(N/A) conjugate acid-base pair is a pair of chemical species that differ by only one proton $(H^{+})$.
When an acid loses a proton,it forms its conjugate base.
When a base gains a proton,it forms its conjugate acid.
The conjugate acid-base pairs for the given species are listed in the table below:

Species	Conjugate Acid/Base
$HNO_{2}$	$NO_{2}^{-}$ (Conjugate Base)
$CN^{-}$	$HCN$ (Conjugate Acid)
$HClO_{4}$	$ClO_{4}^{-}$ (Conjugate Base)
$F^{-}$	$HF$ (Conjugate Acid)
$OH^{-}$	$H_{2}O$ (Conjugate Acid) / $O^{2-}$ (Conjugate Base)
$CO_{3}^{2-}$	$HCO_{3}^{-}$ (Conjugate Acid)
$S^{2-}$	$HS^{-}$ (Conjugate Acid)

(B) Lewis acids are defined as chemical species that can accept a lone pair of electrons.
$BF_3$ has an incomplete octet on the Boron atom,making it an electron-deficient species.
$H^{+}$ is a proton and has an empty $1s$ orbital,allowing it to accept an electron pair.
$NH_4^{+}$ contains a nitrogen atom that has already donated its lone pair to a proton,but the species as a whole can act as a Lewis acid in certain contexts due to the positive charge and the ability to accept electrons to revert to $NH_3$ and $H^{+}$.
$H_2O$ acts as a Lewis base because it has lone pairs on the oxygen atom that it can donate.
Therefore,$BF_3, H^{+},$ and $NH_4^{+}$ are Lewis acids.

The conjugate base of a Bronsted acid is formed by the removal of a proton $(H^+)$ from the acid.

Bronsted acid	Conjugate base
$HF$	$F^-$
$H_2SO_4$	$HSO_4^-$
$HCO_3^-$	$CO_3^{2-}$

The conjugate acid of a Bronsted base is formed by adding a proton $(H^{+})$ to the base. The table below lists the conjugate acids for the given Bronsted bases:

Bronsted base	Conjugate acid
$NH_{2}^{-}$	$NH_{3}$
$NH_{3}$	$NH_{4}^{+}$
$HCOO^{-}$	$HCOOH$

(N/A) Bronsted acid is a proton $(H^{+})$ donor,and a Bronsted base is a proton $(H^{+})$ acceptor. The conjugate acid is formed by adding a proton $(H^{+})$ to the species,while the conjugate base is formed by removing a proton $(H^{+})$ from the species.

Species	Conjugate Acid $(+H^{+})$	Conjugate Base $(-H^{+})$
$H_{2}O$	$H_{3}O^{+}$	$OH^{-}$
$HCO_{3}^{-}$	$H_{2}CO_{3}$	$CO_{3}^{2-}$
$HSO_{4}^{-}$	$H_{2}SO_{4}$	$SO_{4}^{2-}$
$NH_{3}$	$NH_{4}^{+}$	$NH_{2}^{-}$

(N/A) $(a) OH^{-}$ is a Lewis base because it has lone pairs of electrons that it can donate.
$(b) F^{-}$ is a Lewis base because it has lone pairs of electrons that it can donate.
$(c) H^{+}$ is a Lewis acid because it has an empty orbital and can accept a pair of electrons.
$(d) BCl_{3}$ is a Lewis acid because the boron atom has an incomplete octet and can accept a pair of electrons.

The relationship between the ionization constant of an acid $(K_a)$ and its conjugate base $(K_b)$ is given by the equation: $K_a \times K_b = K_w$,where $K_w = 10^{-14}$ at $298 \ K$.
$1$. For $HF$ $(K_a = 6.8 \times 10^{-4})$,the conjugate base is $F^-$.
$K_b(F^-) = \frac{10^{-14}}{6.8 \times 10^{-4}} = 1.47 \times 10^{-11}$.
$2$. For $HCOOH$ $(K_a = 1.8 \times 10^{-4})$,the conjugate base is $HCOO^-$.
$K_b(HCOO^-) = \frac{10^{-14}}{1.8 \times 10^{-4}} = 5.56 \times 10^{-11}$.
$3$. For $HCN$ $(K_a = 4.8 \times 10^{-9})$,the conjugate base is $CN^-$.
$K_b(CN^-) = \frac{10^{-14}}{4.8 \times 10^{-9}} = 2.08 \times 10^{-6}$.