Water hydrolysis and pH scale Questions in English

(D) For a neutral solution,$[H^{+}] = [OH^{-}]$.
Given $pK_w = 13.26$,we know that $K_w = [H^{+}] [OH^{-}] = 10^{-pK_w} = 10^{-13.26}$.
Since $[H^{+}] = [OH^{-}]$,we have $[H^{+}]^2 = 10^{-13.26}$.
Taking the square root,$[H^{+}] = 10^{-13.26 / 2} = 10^{-6.63}$.
By definition,$pH = -\log[H^{+}] = -\log(10^{-6.63}) = 6.63$.
Therefore,the correct option is $(d)$.

(D) When $NaCl$ dissolves in water,it dissociates into $Na^+$ and $Cl^-$ ions.
These ions interact with water molecules through ion-dipole interactions.
This process of surrounding the ions with water molecules is known as hydration.
Therefore,the sodium ion becomes hydrated.

(B) At $298 \ K$,the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$,so $pH = -\log[H^+] = 7.0$.
For heavy water $(D_2O)$,the dissociation constant $(K_w')$ is approximately $1.6 \times 10^{-15}$ at $298 \ K$.
Since $pD = -\log[D^+]$,and $[D^+] = \sqrt{K_w'} = \sqrt{1.6 \times 10^{-15}} \approx 4.0 \times 10^{-8} \ M$.
Therefore,$pD = -\log(4.0 \times 10^{-8}) \approx 7.4$.
Thus,the $pH$ (or $pD$) of $D_2O$ is approximately $7.35$ and for $H_2O$ it is $7.0$.

(C) At room temperature $(25^{\circ}C)$,the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.
For neutral water,$[H^+] = [OH^-] = \sqrt{K_w} = 1.0 \times 10^{-7} \ M$.
The $pH$ is calculated as $pH = -\log[H^+] = -\log(10^{-7}) = 7$.
Therefore,the correct option is $(C)$.

(B) The environment in the human stomach is strongly acidic due to the secretion of hydrochloric acid $(HCl)$.
Therefore,the $pH$ of the stomach is approximately $2$.

(A) At $25\,^oC$,the ionic product of water $(K_w)$ is $10^{-14}$,so $pH = -\log[H^+] = 7$ for a neutral solution.
However,the dissociation of water is an endothermic process.
As the temperature increases to $90\,^oC$,the value of $K_w$ increases (it becomes greater than $10^{-14}$).
Since $K_w = [H^+][OH^-]$,for a neutral solution $[H^+] = [OH^-] = \sqrt{K_w}$.
At $90\,^oC$,$\sqrt{K_w} > 10^{-7}$,which means $[H^+] > 10^{-7}$.
Therefore,$pH = -\log[H^+] < 7$.
Since $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,it does not undergo hydrolysis and the solution remains neutral.
Thus,at $90\,^oC$,the $pH$ of a neutral $NaCl$ solution is less than $7$.

(A) The ionic product of water $(K_w)$ is defined as the product of the concentrations of hydronium ions $(H_3O^+)$ and hydroxide ions $(OH^-)$ in pure water or aqueous solutions.
At $25 \, ^\circ C$ $(298 \, K)$,the concentration of both $H_3O^+$ and $OH^-$ ions is $1.0 \times 10^{-7} \, M$.
Therefore,$K_w = [H_3O^+][OH^-] = (1.0 \times 10^{-7}) \times (1.0 \times 10^{-7}) = 1.0 \times 10^{-14}$.

(B) The ionic product of water $(K_w)$ is related to the dissociation constant $(K)$ and the molar concentration of water $[H_2O]$ by the equation:
$K_w = K \times [H_2O]$
Given $K = 1.8 \times 10^{-16}$ and the molar concentration of water $[H_2O] = 55.5 \ mol \ L^{-1}$.
$K_w = 1.8 \times 10^{-16} \times 55.5 \approx 1 \times 10^{-14}$.

(C) The human body maintains a physiological $pH$ balance through various buffer systems. The $pH$ of the blood and intracellular fluids is typically maintained around $6.8$ to $7.4$ depending on the specific compartment,but in the context of general biological equilibrium questions,$6.8$ is often cited as the physiological neutral $pH$ for intracellular environments.

(D) The auto-ionization of water is an endothermic process: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium to the right.
This results in an increase in the concentration of $H^+$ and $OH^-$ ions.
Since the ionic product of water $K_w = [H^+][OH^-]$,an increase in the concentration of these ions leads to an increase in the value of $K_w$.

(B) Given: $[H^+] = 6 \times 10^{-4} \ M$.
The formula for $pH$ is $pH = -\log[H^+]$.
$pH = -\log(6 \times 10^{-4})$.
$pH = -(\log 6 + \log 10^{-4})$.
$pH = -(\log 6 - 4)$.
$pH = 4 - \log 6$.
Since $\log 6 \approx 0.778$,
$pH = 4 - 0.778 = 3.222 \approx 3.22$.

(A) The ionic product of water $(K_W)$ is defined as the product of the concentrations of hydronium ions and hydroxide ions in water at a given temperature.
$K_W = [H_3O^+][OH^-]$
Since the concentration of both ions is expressed in molarity $(mol \ L^{-1})$,
$K_W = (mol \ L^{-1}) \times (mol \ L^{-1}) = mol^2 \ L^{-2}$
Therefore,the unit of $K_W$ is $mol^2 \ L^{-2}$.

(C) Given: $[OH^{-}] = 10^{-9} \ M$ at $25^{\circ}C$.
We know that $K_w = [H^{+}] [OH^{-}] = 10^{-14}$.
Therefore,$[H^{+}] = \frac{10^{-14}}{10^{-9}} = 10^{-5} \ M$.
The $pH$ is calculated as $pH = -\log[H^{+}]$.
$pH = -\log(10^{-5}) = 5$.

(C) The formula for $pH$ is given by: $pH = -\log[H^+]$
Given $pH = 5.4$,we have: $5.4 = -\log[H^+]$
Therefore,$\log[H^+] = -5.4$
To find $[H^+]$,we take the antilog: $[H^+] = \text{antilog}(-5.4)$
$[H^+] = \text{antilog}(-6 + 0.6)$
$[H^+] = 10^{-6} \times \text{antilog}(0.6)$
Since $\text{antilog}(0.6) \approx 3.98$,we get:
$[H^+] = 3.98 \times 10^{-6} \ mol/L$

(A) The ionic product of water,denoted as $K_w$,is the product of the molar concentrations of hydronium ions $(H_3O^+)$ and hydroxide ions $(OH^-)$ in water at a given temperature.
For pure water,the dissociation reaction is $2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$.
The equilibrium constant expression is $K_c = \frac{[H_3O^+][OH^-]}{[H_2O]^2}$.
Since the concentration of water is considered constant,the ionic product is defined as $K_w = [H_3O^+][OH^-]$.

(A) The formula for $pOH$ is given by: $pOH = -\log[OH^{-}]$
Substituting the given concentration $[OH^{-}] = 10^{-9} \ M$:
$pOH = -\log(10^{-9})$
Using the logarithmic property $\log(a^b) = b \log(a)$:
$pOH = -(-9) \log(10)$
Since $\log(10) = 1$:
$pOH = 9$

(C) The $pH$ is defined as $pH = -\log[H^{+}]$.
For $pH_1 = 3$,$[H^{+}]_1 = 10^{-3} \ M$.
For $pH_2 = 6$,$[H^{+}]_2 = 10^{-6} \ M$.
The ratio of concentrations is $\frac{[H^{+}]_1}{[H^{+}]_2} = \frac{10^{-3}}{10^{-6}} = 10^{3} = 1000$.
Therefore,the $H^{+}$ ion concentration decreases by $1000$ times.

(C) The formula for $pH$ is given by: $pH = - \log [H^{+}]$
Given that $pH = 2$,we substitute the value into the equation:
$2 = - \log [H^{+}]$
Multiply both sides by $-1$:
$-2 = \log [H^{+}]$
Taking the antilog on both sides:
$[H^{+}] = 10^{-2} \ M$

(D) The ionic product of water $(K_w)$ is given by $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$ at $298 \ K$.
The ionization constant of water $(K_a)$ is given by $K_a = \frac{[H^+][OH^-]}{[H_2O]}$.
Since the concentration of water $[H_2O] \approx 55.5 \ M$,we have $K_a = \frac{K_w}{55.5} = \frac{1.0 \times 10^{-14}}{55.5} \approx 1.8 \times 10^{-16}$.
Thus,$K_w$ is much larger than $K_a$.
For option $D$,the number of $H^+$ ions in $1 \ L$ of water is calculated as: $n = [H^+] \times N_A = 10^{-7} \ mol/L \times 6.023 \times 10^{23} \ ions/mol = 6.023 \times 10^{16} \ ions$.
Given the slight variation in constants,$6.23 \times 10^{16}$ is the accepted value in many textbooks.

(D) The $pH$ of a solution is defined as the negative logarithm of the hydronium ion concentration,i.e.,$pH = -\log [H_3O^{+}]$.
Since $-\log [H_3O^{+}] = \log \frac{1}{[H_3O^{+}]}$,option $B$ is also correct.
By rearranging $pH = -\log [H^{+}]$,we get $[H^{+}] = 10^{-pH}$,which is also a valid representation of the relationship.
Therefore,all the given equations represent the $pH$ of a solution correctly.

(C) Given $pH = 3.3$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Therefore,$pOH = 14 - 3.3 = 10.7$.
The concentration of $(OH^-)$ is given by $[OH^-] = 10^{-pOH} = 10^{-10.7}$.
$[OH^-] = 10^{0.3} \times 10^{-11}$.
Since $10^{0.3} \approx 1.995 \approx 2$,we get $[OH^-] \approx 2 \times 10^{-11} \ M$.

(A) The dissociation constant $K$ for water is given by the expression $K = \frac{[H^+][OH^-]}{[H_2O]}$.
At $25^{\circ}C$,the ionic product of water $K_w = [H^+][OH^-] = 10^{-14}$.
The molar concentration of water $[H_2O]$ is calculated as $\frac{1000 \ g/L}{18 \ g/mol} \approx 55.4 \ mol/L$.
Therefore,$K = \frac{10^{-14}}{55.4} = 10^{-14} \times (55.4)^{-1}$.

(B, D) Given: $pOH = 10.0$.
We know that $[OH^-] = 10^{-pOH} = 10^{-10} \ M$.
Using the ionic product of water,$K_w = [H^+][OH^-] = 10^{-14}$.
Therefore,$[H^+] = \frac{K_w}{[OH^-]} = \frac{K_w}{10^{-10}}$.
Also,$[H^+] = 10^{-pH}$. Since $pH + pOH = 14$,$pH = 14 - 10 = 4$.
Thus,$[H^+] = 10^{-4} \ M$.
Both options $(b)$ and $(d)$ are correct.

(B) In an aqueous solution at $298 \ K$,the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.
We know that $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$.
Taking the negative logarithm on both sides,we get $-\log K_w = -\log([H^+][OH^-]) = -\log(1.0 \times 10^{-14})$.
This simplifies to $pK_w = pH + pOH = 14$.

(B) The concentration of hydrogen ions $[H^+]$ is related to $pH$ by the formula: $[H^+] = 10^{-pH}$.
Initially,$pH_1 = 5$,so $[H^+]_1 = 10^{-5} \ M$.
Finally,$pH_2 = 2$,so $[H^+]_2 = 10^{-2} \ M$.
The factor by which the concentration increases is given by the ratio: $\frac{[H^+]_2}{[H^+]_1} = \frac{10^{-2}}{10^{-5}} = 10^{(-2 - (-5))} = 10^3 = 1000$.
Therefore,the hydrogen ion concentration increases by a factor of $1000$.

(B) Given: $pH = 3.7$
The formula for $pH$ is $pH = -\log[H^+]$.
Therefore,$[H^+] = 10^{-pH} = 10^{-3.7}$.
$[H^+] = 10^{0.3} \times 10^{-4}$.
Since $10^{0.3} \approx 1.995 \approx 2.0$,
$[H^+] \approx 2 \times 10^{-4} \, \text{mol/L}$.

(B) The $pH$ scale ranges from $0$ to $14$.
An aqueous solution is acidic if its $pH < 7$.
Since the given $pH$ is $0$,which is less than $7$,the solution is highly acidic.

(D) The auto-ionization of water is represented by the equation: $H_2O(l) \rightleftharpoons H^{+}(aq) + OH^{-}(aq)$.
The equilibrium constant for this reaction is $K_c = \frac{[H^{+}][OH^{-}]}{[H_2O]}$.
Since the concentration of water $[H_2O]$ is essentially constant in dilute aqueous solutions,we define the ionic product of water as $K_w = K_c [H_2O] = [H^{+}][OH^{-}]$.
Therefore,the ionic product of water is the product of the concentrations of hydrogen ions and hydroxide ions.

(A) For an acid $HA$,the dissociation equilibrium is: $HA \rightleftharpoons H^{+} + A^{-}$.
The acid dissociation constant is given by: $K_a = \frac{[H^{+}][A^{-}]}{[HA]}$.
Therefore,the inverse of the acid dissociation constant is: $K_a^{-1} = \frac{1}{K_a} = \frac{[HA]}{[H^{+}][A^{-}]}$.

(B) Pure water has a $pH$ of $7$ at $298 \ K$.
When it absorbs $CO_2$ from the atmosphere,it reacts with water to form carbonic acid $(H_2CO_3)$.
$CO_2 + H_2O \rightleftharpoons H_2CO_3$.
$H_2CO_3$ is a weak acid that dissociates to release $H^+$ ions: $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$.
The increase in $H^+$ ion concentration causes the $pH$ of the solution to decrease below $7$.

(A) Given that $pOH = 13$ at $298 \, K$.
We know that $pH + pOH = 14$.
Therefore,$pH = 14 - 13 = 1$.
$A$ solution with $pH = 1$ is highly acidic because it is much lower than $7$.

(D) The ionic product of water,$K_w$,is temperature-dependent.
At $298 \ K$,$K_w = 1.0 \times 10^{-14}$,so $\sqrt{K_w} = 10^{-7} \ M$.
However,as temperature increases,$K_w$ increases,and the concentration of $[H^{+}]$ and $[OH^{-}]$ in a neutral solution increases above $10^{-7} \ M$.
Therefore,the statement that $[H^{+}] = [OH^{-}] = 10^{-7} \ M$ for a neutral solution at every temperature is incorrect.

(B) In pure water,the dissociation is represented as: $2H_2O \rightleftharpoons H_3O^+ + OH^-$.
Since the water is pure,$[H_3O^+] = [OH^-]$.
Given that $[H_3O^+] = 10^{-6} \, mol \, L^{-1}$,it follows that $[OH^-] = 10^{-6} \, mol \, L^{-1}$.
The ionic product of water is defined as $K_W = [H_3O^+][OH^-]$.
Substituting the values: $K_W = (10^{-6}) \times (10^{-6}) = 10^{-12}$.

(D) When $NaCl$ is dissolved in water,the $Na^+$ ions are surrounded by water molecules,a process known as hydration.

(B) Electrical conductivity in a liquid requires the presence of mobile ions.
Pure water is a very weak electrolyte and undergoes self-ionization to a negligible extent,represented by the equation: $2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$.
Because the concentration of these ions is extremely low ($10^{-7} \ M$ at $298 \ K$),pure water is considered a non-conductor of electricity.

(B) During a thunderstorm,atmospheric $N_2$ and $O_2$ react to form nitrogen oxides $(NO_x)$,which dissolve in rainwater to form nitric acid $(HNO_3)$.
This increases the concentration of $H^+$ ions in the water,thereby lowering the $pH$ value compared to normal rainwater.

(C) Water with a $pH$ less than $6.5$ is considered acidic and is known to cause corrosion in metal water pipes. This acidity increases the solubility of metals,leading to the degradation of the pipe material.

(A) The $pH$ of a solution is calculated using the formula: $pH = -\log [H^{+}]$.
Given,$[H^{+}] = 5.5 \times 10^{-3} \ mol \ L^{-1}$.
Substituting the value into the formula:
$pH = -\log (5.5 \times 10^{-3})$
$pH = -(\log 5.5 + \log 10^{-3})$
$pH = -(\log 5.5 - 3)$
$pH = 3 - \log 5.5$
Since $\log 5.5 \approx 0.74$,
$pH = 3 - 0.74 = 2.26$.

(B) Given: $pH = 4.7$ and $K_w = 10^{-14} \ mol^2/L^2$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Therefore,$pOH = 14 - 4.7 = 9.3$.
$[OH^{-}] = 10^{-pOH} = 10^{-9.3}$.
$[OH^{-}] = 10^{0.7} \times 10^{-10}$.
Since $10^{0.7} \approx 5$,the concentration is $[OH^{-}] = 5 \times 10^{-10} \ M$.

(C) In pure water,the dissociation is represented as $2H_2O \rightleftharpoons H_3O^{+} + OH^{-}$.
Since the water is pure,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}] = 10^{-6}\,mol\,L^{-1}$.
The ionic product of water $(K_w)$ is defined as $K_w = [H_3O^{+}][OH^{-}]$.
Substituting the given values: $K_w = (10^{-6}) \times (10^{-6}) = 10^{-12}$.

(D) At $25\, ^oC$,$K_w = 10^{-14}$.
At $100\, ^oC$,$K_w = 55 \times 10^{-14}$.
For a neutral solution,$[H^+] = [OH^-] = \sqrt{K_w}$.
$[H^+] = \sqrt{55 \times 10^{-14}} = \sqrt{55} \times 10^{-7}$.
$pH = -\log [H^+] = -\log (\sqrt{55} \times 10^{-7}) = -[\frac{1}{2} \log 55 - 7]$.
$pH = 7 - \frac{1}{2} \log 55 = 7 - \frac{1.74}{2} = 7 - 0.87 = 6.13$.

(A) Given the concentration of hydronium ions: $[H_3O^+] = 1 \times 10^{-10} \, M$.
First,calculate the $pH$ of the solution:
$pH = -\log[H_3O^+] = -\log[10^{-10}] = 10$.
At $25 \, ^\circ C$,the relationship between $pH$ and $pOH$ is:
$pH + pOH = 14$.
Substituting the value of $pH$:
$10 + pOH = 14$.
Therefore,$pOH = 14 - 10 = 4$.

(B) The molar conductivity of water at infinite dilution is given by $\lambda_{H_2O}^\infty = \lambda_{H^{+}}^\infty + \lambda_{OH^{-}}^\infty = 350 + 200 = 550 \ S \ cm^2 \ eq^{-1}$.
The relationship between conductivity $\kappa$,molar conductivity $\lambda$,and concentration $C$ is $\lambda = \frac{\kappa \times 1000}{C}$.
Substituting the values: $550 = \frac{5.5 \times 10^{-7} \times 1000}{C_{H^{+}}}$.
Solving for $C_{H^{+}}$: $C_{H^{+}} = \frac{5.5 \times 10^{-4}}{550} = 10^{-6} \ M$.
Since $pH = -\log[H^{+}]$,we get $pH = -\log(10^{-6}) = 6$.

(B) The degree of hydrolysis $h$ is given by the formula: $h = \frac{\lambda_{c} - \lambda_{c'}}{\lambda_{\infty} - \lambda_{c'}}$
Substituting the given values: $h = \frac{122.5 - 106.5}{426.0 - 106.5} = \frac{16}{319.5} \approx 0.05008$
The hydrolysis constant $K_h$ for a salt of a weak base and strong acid is given by: $K_h = \frac{ch^2}{1-h}$
Substituting $c = 0.01 \, M$ and $h = 0.05$: $K_h = \frac{0.01 \times (0.05)^2}{1 - 0.05} = \frac{0.01 \times 0.0025}{0.95} = \frac{2.5 \times 10^{-5}}{0.95} \approx 2.63 \times 10^{-5}$

(D) The ionic product of water is defined as $K_w = [H^+][OH^-]$.
By definition,$pK_w = -\log(K_w)$.
Given $K_w = 10^{-13} \ M^2$,we have $pK_w = -\log(10^{-13}) = 13$.
Since $pH = -\log[H^+]$ and $pOH = -\log[OH^-]$,the relationship between $pH$,$pOH$,and $pK_w$ is given by $pH + pOH = pK_w$.
Therefore,at $62 \ ^\circ C$,the sum $pH + pOH = 13$.

(C) Given $pH = 4$,so $pOH = 14 - 4 = 10$.
The concentration of $OH^{-}$ ions is $[OH^{-}] = 10^{-pOH} = 10^{-10} \ M$.
Since $1 \ M = 1 \ mol/L = 1 \ mol/1000 \ mL$,the number of moles of $OH^{-}$ in $1 \ mL$ is $\frac{10^{-10}}{1000} = 10^{-13} \ mol$.
The number of $OH^{-}$ ions is calculated by multiplying the number of moles by Avogadro's number $(N_A = 6.023 \times 10^{23} \ mol^{-1})$:
$\text{Number of ions} = 10^{-13} \times 6.023 \times 10^{23} = 6.023 \times 10^{10}$.

(A) The auto-ionization of water is an endothermic process: $H_2O(l) \rightleftharpoons H^{\oplus}(aq) + OH^{-}(aq)$.
According to Le Chatelier's principle,as temperature $(T)$ increases,the equilibrium shifts to the right.
This leads to an increase in the concentration of both $H^{\oplus}$ and $OH^{-}$ ions.
Since $pOH = -\log[OH^{-}]$,an increase in $[OH^{-}]$ results in a decrease in $pOH$.

(D) For a neutral solution at $60\,^{\circ}C$,$[H_3O^{+}] = [OH^{-}]$.
Since $K_w = [H_3O^{+}][OH^{-}] = 9.6 \times 10^{-14}$,we have $[H_3O^{+}]^2 = 9.6 \times 10^{-14}$.
Therefore,$[H_3O^{+}] = \sqrt{9.6 \times 10^{-14}} \approx 3.1 \times 10^{-7} \,M$.
The neutral $pH$ at $60\,^{\circ}C$ is $pH = -\log(3.1 \times 10^{-7}) \approx 6.51$.
Since the neutral $pH$ is $6.51$,a solution with $pH = 7.0$ (which is greater than $6.51$) is basic in nature.

(C) Given $pK_w = 12.60$,so $K_w = [H^{+}][OH^{-}] = 10^{-12.60}$.
For a neutral solution,$[H^{+}] = [OH^{-}]$.
Thus,$[H^{+}]^2 = 10^{-12.60} = 10^{0.40} \times 10^{-13} \approx 2.51 \times 10^{-13} \approx 0.251 \times 10^{-12}$.
Calculating $[H^{+}] = \sqrt{10^{-12.60}} = 10^{-6.30} \approx 5.01 \times 10^{-7} \ M$.
For an acidic solution,$[H^{+}] > 5.01 \times 10^{-7} \ M$.
Since $[H^{+}][OH^{-}] = 10^{-12.60}$,if $[H^{+}] > 5.01 \times 10^{-7} \ M$,then $[OH^{-}] < 5.01 \times 10^{-7} \ M$.

(B) At $25\,^{\circ}C$,the dissociation of water is represented as:
$H_2O \rightleftharpoons H^{+} + OH^{-}$
According to the common ion effect,the presence of either $H^{+}$ or $OH^{-}$ ions suppresses the dissociation of water.
At $pOH = 7$,the concentration of $OH^{-}$ is $10^{-7} \, M$ and $[H^{+}]$ is $10^{-7} \, M$. This represents the neutral point where the concentration of both ions is minimal and equal,allowing for the standard equilibrium dissociation.
If $pOH = 14$,the solution is highly basic,$[OH^{-}]$ is very high,and the reaction shifts backward.
If $pOH = 0$,the solution is highly acidic,$[H^{+}]$ is very high,and the reaction shifts backward.
Therefore,the dissociation of water is maximum at $pOH = 7$.