Q: The dissociation of water at \$25^{\circ}\$C is \$ 1.9 \times 10^{-7}\%\$ and the density of water is \$1.0\$ g/cc. The ionisation constant of water is :

The degree of dissociation \$(\alpha)\$ for water is \$\dfrac {1.9 \times 10^{-7}} {100}=1.9 \times 10^{-9}\$.\$1\$ L of water (molecular mass \$= 18\$ g/mol) weighs \$1000\$ g.Hence, concentration of water \$=\dfrac {1000}{18}\$ M.The expression for the ionization constant of water is \$K=\dfrac{[H^+][OH^-]}{[H_2O]}=C (\alpha)^2\$.Substituting values in the above expression, we get:\$K=\dfrac {1000}{18} \times (1.9 \times 10^{-9})^2=2.0 \times 10^{-16}\$.

Q: The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 ml of 1.0 M HA solution is diluted to 32 ml, the degree of dissociation of acid and \$H_{3}O^{+}\$ ion concentration in the resulting solution will be respectively:

Given, \$\alpha =0.5 \%= 0.005\$ and \$C= 0.1\ M\$The expression for the degree of dissociation is \$\alpha = \sqrt {\dfrac{K_a}{C}}\$.\$0.005= \sqrt {\dfrac{K_a}{0.1}} \$\$K_a=2.5 \times 10^{-6}\$.When the solution is diluted, the molarity of the solution is given by the following expression.\$M_1V_1=M_2V_2\$or \$2 \times 1 = 32 \times M_2\$Hence, \$M_2=\dfrac {1} {16} \$.For this diluted solution, \$\alpha = \sqrt {\dfrac{K_a}{C}}=0.632\%\$.The hydrogen ion concentration is \$[H^+]=c \alpha =\dfrac {1} {16} \times 0.00632 =3.955 \times 10^{-4}\$.

Q: Which of the following statements is correct?

\$HCl\$ in water always creates an acidic solution. A solution with pH 8 is basic but the pH of the solution with \$10^{-8} \$ molarity \$HCl\$ must be less than 7.Species formed after protonation of is called a conjugate base. The conjugate base of \$H_2PO_4^{-2}\$ is \$HPO_4^{-3}\$.Autoprotolysis is the dissociation of water which increases with increase in temperature.Half neutralization point occurs when half of the acid moles are consumed by strong base. Therefore, \$pH = 1/2 \left(pK_a - log (c/2) \right) \$.

Q: Total number of moles for the reaction \$2HI\$ \$\rightleftharpoons\$ \$H_{2}+I_{2}\$, if \$\alpha\$ is degree of dissociation is :

\$ HI\$ \$\rightleftharpoons\$ \$H_2\$ + \$I_2\$Initial moles 2 0 0Change \$-\alpha\$ \$\dfrac {\alpha} {2}\$\$\dfrac {\alpha} {2}\$moles at equilibrium \$2-\alpha\$ \$\dfrac {\alpha} {2}\$\$\dfrac {\alpha} {2}\$The reaction is \$2HI \rightleftharpoons H_{2}+I_{2}\$.\$\alpha\$ is the degree of dissociation.Total number of moles \$=2-\alpha+\frac {\alpha} {2}+\frac {\alpha} {2}=2\$.

Q: \$H_{2}O\$ is stronger acid than \$D_{2}O\$.

At \$25^{\circ}C\$ ionic product of water is \$1.0\times 10^{-14}\$ and that of \$D_{2}O\$ is \$1.35\times 10^{-15}\$.Read the above assertion and reason and choose the correct option regarding it.

Question 1

The dissociation of water at $25^{\circ}$C is $ 1.9 \times 10^{-7}\%$ and the density of water is $1.0$ g/cc. The ionisation constant of water is :

Accepted Answer

The degree of dissociation $(\alpha)$ for water is $\dfrac {1.9 \times 10^{-7}} {100}=1.9 \times 10^{-9}$.$1$ L of water (molecular mass $= 18$ g/mol) weighs $1000$ g.Hence, concentration of water $=\dfrac {1000}{18}$ M.The expression for the ionization constant of water is $K=\dfrac{[H^+][OH^-]}{[H_2O]}=C (\alpha)^2$.Substituting values in the above expression, we get:$K=\dfrac {1000}{18} \times (1.9 \times 10^{-9})^2=2.0 \times 10^{-16}$.

Question 2

The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 ml of 1.0 M HA solution is diluted to 32 ml, the degree of dissociation of acid and $H_{3}O^{+}$ ion concentration in the resulting solution will be respectively:

Accepted Answer

Given, $\alpha =0.5 \%= 0.005$ and $C= 0.1\ M$The expression for the degree of dissociation is $\alpha = \sqrt {\dfrac{K_a}{C}}$.$0.005= \sqrt {\dfrac{K_a}{0.1}} $$K_a=2.5 \times 10^{-6}$.When the solution is diluted, the molarity of the solution is given by the following expression.$M_1V_1=M_2V_2$or $2 \times 1 = 32 \times M_2$Hence, $M_2=\dfrac {1} {16} $.For this diluted solution, $\alpha = \sqrt {\dfrac{K_a}{C}}=0.632\%$.The hydrogen ion concentration is $[H^+]=c \alpha =\dfrac {1} {16} \times 0.00632 =3.955 \times 10^{-4}$.

Question 3

Which of the following statements is correct?

Accepted Answer

$HCl$ in water always creates an acidic solution. A solution with pH 8 is basic but the pH of the solution with $10^{-8} $ molarity $HCl$ must be less than 7.Species formed after protonation of is called a conjugate base. The conjugate base of $H_2PO_4^{-2}$ is $HPO_4^{-3}$.Autoprotolysis is the dissociation of water which increases with increase in temperature.Half neutralization point occurs when half of the acid moles are consumed by strong base. Therefore, $pH = 1/2 \left(pK_a - log (c/2) \right) $.

Question 4

Total number of moles for the reaction  $2HI$ $\rightleftharpoons$ $H_{2}+I_{2}$, if $\alpha$  is degree of dissociation is :

Accepted Answer

$ HI$ $\rightleftharpoons$          $H_2$             +  $I_2$Initial moles        2            0                    0Change      $-\alpha$            $\dfrac {\alpha} {2}$$\dfrac {\alpha} {2}$moles at equilibrium         $2-\alpha$            $\dfrac {\alpha} {2}$$\dfrac {\alpha} {2}$The reaction is $2HI \rightleftharpoons H_{2}+I_{2}$.$\alpha$ is the degree of dissociation.Total number of moles $=2-\alpha+\frac {\alpha} {2}+\frac {\alpha} {2}=2$.

Question 5

The first and second dissociation constants of an acid $H_{2}A$ are $1.0\times10^{-5}$ and $ 5.0\times 10^{-10}$ respectively. The overall dissociation constant of the acid will be:

Accepted Answer

For the dissociation of a dibasic acid, the overall dissociation constant is equal to the product of the first and second dissociation constants.Therefore, $K=K_1 \times K_2=1.0 \times 10^{-5} \times 5.0 \times 10^{-10}= 5.0 \times 10^{-15}$Hence, the overall dissociation constant is $5.0\times 10^{-15}$.

Question 6

If the dissociation constant of a weak acid is $1.0 \times 10^{-5}$, then the equilibrium constant for the reaction of the acid with a strong base is:

Accepted Answer

The equilibrium reaction for the dissociation of the weak acid is as follows:$HA \rightleftharpoons H^++A^-$The expression for the dissociation constant $(K_a)$ is as given below:$K_a= \dfrac {[H^+][A^-]} {[HA]}$  $.....(i)$The reaction for the neutralization of weak acid $HA$ with strong base $OH^-$ is as follows:$HA+OH^- \rightleftharpoons A^- + H_2O$The expression for the equilibrium constant is as given below:$K=\dfrac {[A^-]} {[HA][OH^-]}$    $......(ii)$On dividing equation $(i)$ by equation $(ii)$, we get$\dfrac {K_a} {K}=\dfrac {\dfrac {[H^+][A^-]} {[HA]}} {\dfrac {[A^-]} {[HA][OH^-]}}$Hence, $\dfrac {K_a} {K}=[H^+][OH^-]=K_w=10^{-14}$Thus, $K=\dfrac {K_a} {K_w}=\dfrac {10^{-5}} {10^{-14}}=10^9$Hence, the equilibrium constant for the reaction of the acid with a strong base is $1.0\times 10^9$.

Question 7

$H_{2}O$ is stronger acid than $D_{2}O$.

Accepted Answer

At $25^{\circ}C$ ionic product of water is $1.0\times 10^{-14}$ and that of $D_{2}O$ is  $1.35\times 10^{-15}$.Read the above assertion and reason and choose the correct option regarding it.

Question 8

A precipitate of calcium oxalate will not dissolve in:

Accepted Answer

Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of a strong electrolyte, which provides an ion common to that provided by a weak electrolyte.When strong acids such as hydrochloric acid, nitric acid, and aqua regia are added to a solution containing calcium oxalate, the dissociation of oxalic acid is suppressed due to hydrogen ions provided by strong acids (common ion effect). Hence, the concentration of oxalate ions decreases. The precipitate of calcium oxalate will dissolve in it because the product of the concentrations of the ions present in calcium oxalate will be less than the solubility product of calcium oxalate.When acetic acid is added, the common ion effect is not observed, as acetic acid is a weak acid. Hence, the precipitate of calcium oxalate will not dissolve.Option D is the correct answer.

Question 9

Blood $pH$ is controlled by the concentrations of $H_{2}CO_{3}$ and $HCO_{3}{^{-}}$. In the presence of $NaHCO_{3}$, $pH$ of blood:

Accepted Answer

Acidic buffer solution is prepared by mixing a weak acid and its salt with a strong base.Thus, a mixture of $H_{2}CO_{3}$ and $HCO_{3}^{-} $ acts as an acidic buffer solution.The expression for the pH of the acidic buffer solution is as given below:$pH=pK_a+log \frac {[salt]} {[acid]}$In the presence of $NaHCO_{3}$, the concentration of salt increases. Hence, the term $log \frac {[salt]} {[acid]}$ increases. This increases the $pH$ of blood.

Question 10

Which of the following mixture is not a buffer solution?

Accepted Answer

An acidic buffer solution is prepared by mixing a weak acid and its salt with a strong base. A basic buffer solution is prepared by mixing a weak base and its salt with strong acid. A buffer solution cannot be obtained by mixing a strong acid with a strong base.$HCl$ is a strong acid and $NaOH$ is a strong base, therefore it forms salt and not a buffer solution.$NH_4OH$ and $CH_3COOH$ is a mixture of a weak acid and weak base, therefore it forms salt and not buffer solution. Hence option C & D is correct.

Equilibrium

Chemistry
8707 Views

The dissociation of water at $2 5^{\circ}$ C is $1.9 \times 1 0^{- 7} %$ and the density of water is $1.0$ g/cc. The ionisation constant of water is :

The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 ml of 1.0 M HA solution is diluted to 32 ml, the degree of dissociation of acid and $H_{3} O^{+}$ ion concentration in the resulting solution will be respectively:

Which of the following statements is correct?

Total number of moles for the reaction $2 H I$ $⇌$ $H_{2} + I_{2}$ , if $α$ is degree of dissociation is :

The first and second dissociation constants of an acid $H_{2} A$ are $1.0 \times 1 0^{- 5}$ and $5.0 \times 1 0^{- 10}$ respectively. The overall dissociation constant of the acid will be:

If the dissociation constant of a weak acid is $1.0 \times 1 0^{- 5}$ , then the equilibrium constant for the reaction of the acid with a strong base is:

$H_{2} O$ is stronger acid than $D_{2} O$ .

At $2 5^{\circ} C$ ionic product of water is $1.0 \times 1 0^{- 14}$ and that of $D_{2} O$ is $1.35 \times 1 0^{- 15}$ .

Read the above assertion and reason and choose the correct option regarding it.

A precipitate of calcium oxalate will not dissolve in:

Blood $p H$ is controlled by the concentrations of $H_{2} C O_{3}$ and $H C O_{3}^{-}$ . In the presence of $N a H C O_{3}$ , $p H$ of blood:

Which of the following mixture is not a buffer solution?

Equilibrium

Chemistry 8707 Views

The dissociation of water at 25∘C is 1.9×10−7% and the density of water is 1.0 g/cc. The ionisation constant of water is :

The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 ml of 1.0 M HA solution is diluted to 32 ml, the degree of dissociation of acid and H3​O+ ion concentration in the resulting solution will be respectively:

Which of the following statements is correct?

Total number of moles for the reaction 2HI ⇌ H2​+I2​, if α is degree of dissociation is :

The first and second dissociation constants of an acid H2​A are 1.0×10−5 and 5.0×10−10 respectively. The overall dissociation constant of the acid will be:

If the dissociation constant of a weak acid is 1.0×10−5, then the equilibrium constant for the reaction of the acid with a strong base is:

H2​O is stronger acid than D2​O.

At 25∘C ionic product of water is 1.0×10−14 and that of D2​O is 1.35×10−15. Read the above assertion and reason and choose the correct option regarding it.

A precipitate of calcium oxalate will not dissolve in:

Blood pH is controlled by the concentrations of H2​CO3​ and HCO3​−. In the presence of NaHCO3​, pH of blood:

Which of the following mixture is not a buffer solution?

Chemistry
8707 Views

The dissociation of water at $2 5^{\circ}$ C is $1.9 \times 1 0^{- 7} %$ and the density of water is $1.0$ g/cc. The ionisation constant of water is :

The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 ml of 1.0 M HA solution is diluted to 32 ml, the degree of dissociation of acid and $H_{3} O^{+}$ ion concentration in the resulting solution will be respectively:

Total number of moles for the reaction $2 H I$ $⇌$ $H_{2} + I_{2}$ , if $α$ is degree of dissociation is :

The first and second dissociation constants of an acid $H_{2} A$ are $1.0 \times 1 0^{- 5}$ and $5.0 \times 1 0^{- 10}$ respectively. The overall dissociation constant of the acid will be:

If the dissociation constant of a weak acid is $1.0 \times 1 0^{- 5}$ , then the equilibrium constant for the reaction of the acid with a strong base is:

$H_{2} O$ is stronger acid than $D_{2} O$ .

At $2 5^{\circ} C$ ionic product of water is $1.0 \times 1 0^{- 14}$ and that of $D_{2} O$ is $1.35 \times 1 0^{- 15}$ .

Read the above assertion and reason and choose the correct option regarding it.

Blood $p H$ is controlled by the concentrations of $H_{2} C O_{3}$ and $H C O_{3}^{-}$ . In the presence of $N a H C O_{3}$ , $p H$ of blood: