Water hydrolysis and pH scale Questions in English

(B) Given $pH = 13$.
Since $pH = -\log[H^{+}]$,we have $[H^{+}] = 10^{-pH} = 10^{-13} \ M$.
Volume of solution $= 2 \ mL = 2 \times 10^{-3} \ L$.
Moles of $H^{+}$ ions $= \text{Molarity} \times \text{Volume in Litres} = 10^{-13} \times 2 \times 10^{-3} = 2 \times 10^{-16} \ mol$.
Number of $H^{+}$ ions $= \text{Moles} \times N_A = 2 \times 10^{-16} \times 6.022 \times 10^{23} = 12.044 \times 10^7$ ions.

(A) Rain water becomes acidic because atmospheric gases like $CO_2$ dissolve in it to form carbonic acid.
Therefore,the $pH$ of unpolluted rain water is approximately $5.6$.

(B) The ionic product of water is given by the expression: $K_w = [H_3O^{+}][OH^{-}]$.
For pure water,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}] = 10^{-6}\, mol\, L^{-1}$.
Substituting these values into the expression for $K_w$:
$K_w = (10^{-6}) \times (10^{-6}) = 10^{-12}$.

(D) The $pH$ is calculated using the formula: $pH = -\log [H^{+}]$.
Given $[H^{+}] = 2 \times 10^{-9} \ M$.
$pH = -\log (2 \times 10^{-9})$
$pH = -(\log 2 + \log 10^{-9})$
$pH = -(\log 2 - 9)$
$pH = 9 - \log 2$
Since $\log 2 \approx 0.301$,
$pH = 9 - 0.301 = 8.699 \approx 8.7$.

(C) $BiCl_3 + H_2O \to BiOCl + 2HCl$
The reaction involves the hydrolysis of bismuth trichloride in the presence of excess water,resulting in the formation of bismuth oxychloride $(BiOCl)$,which appears as a white precipitate.

(A) The dissociation constant of water $(K_a)$ is related to the ionic product of water $(K_w)$ by the equation: $K_a = \frac{K_w}{[H_2O]}$.
At $30\,^{\circ}C$,the molar concentration of water $[H_2O]$ is approximately $55.5 \, M$ (calculated as $\frac{1000 \, g/L}{18 \, g/mol} \approx 55.5 \, mol/L$).
Therefore,the ratio $\frac{K_a}{K_w} = \frac{1}{[H_2O]} = \frac{1}{55.5}$.
Converting this to a ratio of $18:1000$: $\frac{1}{55.5} \approx 0.018 = \frac{18}{1000}$.

(C) The standard Gibbs energy change is given by the formula $\Delta G^{\circ} = -RT \ln K_w = -2.303 \times RT \times \log K_w$.
At $25\ ^oC$ $(298 \ K)$,the ionic product of water $K_w = 10^{-14}$.
Substituting the values: $\Delta G^{\circ} = -2.303 \times (8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times 298 \ K \times \log(10^{-14})$.
$\Delta G^{\circ} = -2.303 \times 8.314 \times 10^{-3} \times 298 \times (-14) \approx 79.9 \ kJ/mol$.
The closest integer value provided in the options is $90 \ kJ/mol$.

(B) Given that $pOH = 12$ at $25\,^{\circ}C$.
We know that $pH + pOH = 14$.
Therefore,$pH = 14 - pOH = 14 - 12 = 2$.
The concentration of $H^{+}$ ions is given by the formula $[H^{+}] = 10^{-pH}$.
Substituting the value of $pH$,we get $[H^{+}] = 10^{-2} \ M$.

(A) The reaction is the auto-ionization of water: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
The equilibrium constant $K_w$ for this reaction at $25 \, ^\circ C$ is $1.0 \times 10^{-14}$.
The standard Gibbs energy change is given by the formula: $\Delta G^\circ = -RT \ln K_w$.
Substituting the values: $R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$,$T = 298 \, K$,and $K_w = 10^{-14}$.
$\Delta G^\circ = -(8.314 \, J \cdot K^{-1} \cdot mol^{-1}) \times (298 \, K) \times \ln(10^{-14})$.
$\Delta G^\circ = -8.314 \times 298 \times (-14 \times 2.303) \, J/mol$.
$\Delta G^\circ \approx 80400 \, J/mol = 80.4 \, kJ/mol$.
Among the given options,the closest value is $+80 \, kJ$.

(C) The formula for $pH$ is $pH = -\log [H^{+}]$.
When $pH = 5.0$,the hydrogen ion concentration is $[H^{+}]_{1} = 10^{-5} \ M$.
When the $pH$ is decreased to $2.0$,the new hydrogen ion concentration is $[H^{+}]_{2} = 10^{-2} \ M$.
The increase in hydrogen ion concentration is given by the ratio $\frac{[H^{+}]_{2}}{[H^{+}]_{1}} = \frac{10^{-2}}{10^{-5}} = 10^{(-2 - (-5))} = 10^{3} = 1000$.
Therefore,the concentration increases by $1000$ times.

(D) The relationship between $pH$ and hydronium ion concentration is given by the formula: $pH = -\log [H_3O^{+}]$.
Given $pH = 4.30$,we have $4.30 = -\log [H_3O^{+}]$.
Therefore,$[H_3O^{+}] = 10^{-4.30}$.
$[H_3O^{+}] = 10^{0.70} \times 10^{-5} \approx 5.01 \times 10^{-5} \, M$.
Thus,the correct concentration is $5.0 \times 10^{-5} \, M$.

(B) Step $1$: Calculate the millimoles of $HCl$ and $KOH$.
Millimoles of $HCl = 100\, mL \times 0.2\, M = 20\, mmol$.
Millimoles of $KOH = 200\, mL \times 0.1\, M = 20\, mmol$.
Step $2$: Determine the nature of the resulting solution.
Since the millimoles of $HCl$ (acid) equal the millimoles of $KOH$ (base),the reaction $HCl + KOH \rightarrow KCl + H_2O$ results in a neutral solution.
Step $3$: Calculate the $pH$ of the neutral solution at $90\,^oC$.
At $90\,^oC$,the concentration of $[H^{+}]$ in pure water is $10^{-6}\,M$.
For a neutral solution,$[H^{+}] = [OH^{-}] = 10^{-6}\,M$.
$pH = -\log[H^{+}] = -\log(10^{-6}) = 6$.

(A) The dissociation of water is an endothermic process: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
As temperature increases,the equilibrium constant for the dissociation of water $(K_w)$ increases.
At $298 \ K$,$K_w = 10^{-14}$,so $pH = 7$.
At $320 \ K$,$K_w > 10^{-14}$,which means $[H^+]$ and $[OH^-]$ both increase.
Since $pH = -\log[H^+]$,an increase in $[H^+]$ leads to a decrease in $pH$.
Similarly,$pOH = -\log[OH^-]$,so an increase in $[OH^-]$ leads to a decrease in $pOH$.
Therefore,the $pH$ of the neutral solution decreases at higher temperatures.

(B) In an aqueous solution at $25^{\circ}C$,the ionic product of water is $K_w = [H^+][OH^-] = 10^{-14}$.
For an acidic solution,$[H^+] > 10^{-7} \ M$.
Since $[OH^-] = \frac{K_w}{[H^+]}$,if $[H^+] > 10^{-7} \ M$,then $[OH^-] < \frac{10^{-14}}{10^{-7}} \ M$.
Therefore,$[OH^-] < 10^{-7} \ M$.

(D) The initial $pH$ of the solution is $2$. The concentration of hydronium ions $[H_3O^+]$ is given by the formula $[H_3O^+] = 10^{-pH}$.
For $pH = 2$,$[H_3O^+]_{initial} = 10^{-2} \, M$.
If the $pH$ is doubled,the new $pH = 2 \times 2 = 4$.
The new concentration of hydronium ions is $[H_3O^+]_{final} = 10^{-4} \, M$.
The ratio of the final concentration to the initial concentration is $\frac{10^{-4}}{10^{-2}} = 10^{-2} = \frac{1}{100}$.
Therefore,the concentration of hydronium ions decreases to $1/100$th of its original value.

(A) The $p^H$ scale ranges from $0$ to $14$.
$p^H = -\log[H^+]$.
If $p^H = 0$,then $-\log[H^+] = 0$,which implies $[H^+] = 10^0 = 1 \ M$.
$A$ solution with a high concentration of hydrogen ions $([H^+] = 1 \ M)$ is highly acidic.
Therefore,a solution with $p^H = 0$ is acidic.

(A) Given that the $pH$ of the aqueous solution is $6$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Therefore,$pOH = 14 - pH = 14 - 6 = 8$.
Since $pOH = -\log[OH^{-}]$,we have $[OH^{-}] = 10^{-pOH}$.
Substituting the value,$[OH^{-}] = 10^{-8} \ M$.

(A) Given that the $pH$ of the solution is $3$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Therefore,$pOH = 14 - pH = 14 - 3 = 11$.
The concentration of hydroxyl ions $[OH^-]$ is given by the formula $[OH^-] = 10^{-pOH}$.
Substituting the value,$[OH^-] = 10^{-11} \ M$.

(A) For a neutral solution,the concentration of hydrogen ions $[H^+]$ is equal to the concentration of hydroxide ions $[OH^-]$.
Since $K_w = [H^+][OH^-]$,for a neutral solution,$K_w = [H^+]^2$.
Taking the negative logarithm on both sides,we get $pK_w = 2pH$.
Given $pK_w = 13.36$,therefore $pH = \frac{13.36}{2} = 6.68$.

(B) The $pH$ of human blood is slightly basic,typically ranging between $7.35$ and $7.45$. Therefore,it is greater than $7$ but less than $8$.

(A) Given: $pH = 4.70$ and $pK_w = 14$.
We know that $pH + pOH = pK_w$.
Therefore,$pOH = 14 - 4.70 = 9.30$.
The concentration of $OH^-$ ions is given by $[OH^-] = 10^{-pOH} = 10^{-9.30}$.
$[OH^-] = 10^{0.70} \times 10^{-10}$.
Since $10^{0.70} \approx 5.01$,we get $[OH^-] \approx 5 \times 10^{-10} \ M$.

(D) For pure water,the dissociation equilibrium is $H_2O \rightleftharpoons H^+ + OH^-$.
At any temperature,the ionic product of water is $K_w = [H^+][OH^-]$.
Since $[H^+] = [OH^-]$ in pure water,$K_w = [H^+]^2$.
Taking the negative logarithm on both sides,we get $pK_w = -\log(K_w) = -\log([H^+]^2) = -2\log[H^+] = 2pH$.
Therefore,$pH = \frac{pK_w}{2}$.
Given $pK_w = 13.26$ at $50\,^oC$,we have $pH = \frac{13.26}{2} = 6.63$.

(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,increasing the temperature shifts the equilibrium to the right.
Therefore,the concentration of $H^+$ and $OH^-$ ions increases,which leads to an increase in the ionic product of water $(K_w)$.

(D) Given: $pH = 12$ and Volume $V = 10 \ mL = 0.01 \ L$.
Since $pH = -\log[H^{+}]$,we have $[H^{+}] = 10^{-pH} = 10^{-12} \ mol/L$.
The number of moles of $H^{+}$ ions = $[H^{+}] \times V = 10^{-12} \ mol/L \times 0.01 \ L = 10^{-14} \ mol$.
The number of $H^{+}$ ions = $\text{moles} \times N_A = 10^{-14} \times 6.023 \times 10^{23} = 6.023 \times 10^{9}$.

(D) When $NaCl$ is dissolved in water,it dissociates into $Na^+$ and $Cl^-$ ions.
The $Na^+$ ions are surrounded by water molecules due to ion-dipole interactions,which is known as hydration.
Therefore,the sodium ion undergoes hydration.

(B) The dissociation of water is represented as: $H_{2}O(\ell) \rightleftharpoons H_{(aq)}^{+} + OH_{(aq)}^{-}$.
The ionization of water is an endothermic process,meaning $\Delta H > 0$.
According to Le Chatelier's principle,increasing the temperature shifts the equilibrium in the forward direction,which increases the concentration of both $H^{+}$ and $OH^{-}$ ions.
Since $pH = -\log[H^{+}]$,an increase in $[H^{+}]$ leads to a decrease in $pH$.
Therefore,the assertion is false because the $pH$ of water decreases with an increase in temperature,and the reason is false because the dissociation of water is endothermic,not exothermic.

(N/A) Hydrolysis is defined as a chemical reaction in which hydrogen and hydroxide ions ($H^{+}$ and $OH^{-}$ ions) of a water molecule react with a compound to form new products. For example:
$NaH + H_2O \longrightarrow NaOH + H_2$
Hydration is defined as the addition of one or more water molecules to ions or molecules to form hydrated compounds,where no chemical bond breaking of the water molecule occurs. For example:
$CuSO_4 + 5H_2O \longrightarrow CuSO_4 \cdot 5H_2O$

(A) Given:
$\left[ H^{+} \right] = 3.8 \times 10^{-3} \ M$
The formula for $pH$ is:
$pH = -\log \left[ H^{+} \right]$
Substituting the value:
$pH = -\log \left( 3.8 \times 10^{-3} \right)$
$pH = -(\log 3.8 + \log 10^{-3})$
$pH = -(\log 3.8 - 3)$
$pH = 3 - \log 3.8$
Using $\log 3.8 \approx 0.58$:
$pH = 3 - 0.58 = 2.42$
Thus,the $pH$ of the soft drink is $2.42$.

(N/A) Given,$pH = 3.76$.
It is known that,
$pH = -\log [H^{+}]$
$\Rightarrow \log [H^{+}] = -pH$
$\Rightarrow [H^{+}] = \text{antilog}(-pH)$
$= \text{antilog}(-3.76)$
$= 1.74 \times 10^{-4} \ M$.
Hence,the concentration of hydrogen ion in the given sample of vinegar is $1.74 \times 10^{-4} \ M$.

The concentration of hydrogen ions $[H^{+}]$ is calculated using the formula: $pH = -\log [H^{+}]$,which implies $[H^{+}] = 10^{-pH}$.
$(a)$ Human muscle-fluid,$pH = 6.83$
$[H^{+}] = 10^{-6.83} = 1.48 \times 10^{-7} \ M$
$(b)$ Human stomach fluid,$pH = 1.2$
$[H^{+}] = 10^{-1.2} = 6.31 \times 10^{-2} \ M = 0.063 \ M$
$(c)$ Human blood,$pH = 7.38$
$[H^{+}] = 10^{-7.38} = 4.17 \times 10^{-8} \ M$
$(d)$ Human saliva,$pH = 6.4$
$[H^{+}] = 10^{-6.4} = 3.98 \times 10^{-7} \ M$

The hydrogen ion concentration $[H^+]$ is calculated using the formula: $pH = -\log [H^+]$,which implies $[H^+] = 10^{-pH}$.
$(i)$ For milk $(pH = 6.8)$: $[H^+] = 10^{-6.8} = 1.58 \times 10^{-7} \, M$
$(ii)$ For black coffee $(pH = 5.0)$: $[H^+] = 10^{-5.0} = 1.0 \times 10^{-5} \, M$
$(iii)$ For tomato juice $(pH = 4.2)$: $[H^+] = 10^{-4.2} = 6.31 \times 10^{-5} \, M$
$(iv)$ For lemon juice $(pH = 2.2)$: $[H^+] = 10^{-2.2} = 6.31 \times 10^{-3} \, M$
$(v)$ For egg white $(pH = 7.8)$: $[H^+] = 10^{-7.8} = 1.58 \times 10^{-8} \, M$

(A) Ionic product,$K_{w} = [H^{+}] [OH^{-}]$.
Let $[H^{+}] = x$.
Since $[H^{+}] = [OH^{-}]$ for neutral water,$K_{w} = x^{2}$.
Given $K_{w}$ at $310 \, K$ is $2.7 \times 10^{-14}$.
Therefore,$2.7 \times 10^{-14} = x^{2}$.
$x = \sqrt{2.7 \times 10^{-14}} = 1.64 \times 10^{-7}$.
So,$[H^{+}] = 1.64 \times 10^{-7} \, M$.
$pH = -\log [H^{+}] = -\log [1.64 \times 10^{-7}]$.
$pH = 7 - \log(1.64) = 7 - 0.215 = 6.785 \approx 6.78$.
Hence,the $pH$ of neutral water at $310 \, K$ is $6.78$.

Water is unique in its ability to act as both an acid and a base. In pure water,one $H_2O$ molecule donates a proton and acts as an acid,while another accepts a proton and acts as a base. The following equilibrium exists:
$H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons H_3O^{+}_{(aq)} + OH^{-}_{(aq)}$
The equilibrium constant $K$ is given by:
$K = \frac{[H_3O^{+}][OH^{-}]}{[H_2O]^2}$
Since the concentration of water $[H_2O]$ is constant,it is incorporated into the equilibrium constant to define a new constant,$K_w$,known as the ionic product of water:
$K_w = K[H_2O]^2 = [H_3O^{+}][OH^{-}]$
At $298 \ K$,the concentration of $H_3O^{+}$ and $OH^{-}$ is $1.0 \times 10^{-7} \ M$. Therefore:
$K_w = (1.0 \times 10^{-7})(1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \ M^2$

(N/A) Calculation of the concentration of pure water:
Density of pure water $= 1.0 \ g \ mL^{-1} = 1000 \ g \ L^{-1}$.
Concentration of water $= \frac{\text{Mass of } 1 \ L \text{ water}}{\text{Molar mass of } H_2O} = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol \ L^{-1}$.
The dissociation equilibrium of water is $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
The ionic product of water at $298 \ K$ is $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$.
In pure water,$[H^+] = [OH^-] = 1.0 \times 10^{-7} \ M$.
The ratio of dissociated water to undissociated water is $\frac{1.0 \times 10^{-7}}{55.55} \approx 1.8 \times 10^{-9}$.
Since this value is extremely small,the equilibrium lies heavily towards the left side (undissociated water).

(N/A) $pH$ Scale: The hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the $pH$ scale.
Definition of $pH$: The $pH$ of a solution is defined as the negative logarithm of the activity of hydrogen ions $(a_{H^{+}})$.
Since activity $a$ is dimensionless,$a_{H^{+}} = [H^{+}] / (1 \ mol \ L^{-1})$.
This definition is expressed as:
$pH = -\log a_{H^{+}} \quad \dots (Eq.-I)$
For dilute solutions where the concentration is expressed in $mol \ L^{-1}$:
$pH = -\log [H^{+}] \quad \dots (Eq.-II)$
The $pH$ scale is suitable for dilute solutions where $[H^{+}] < 1 \ M$.
Temperature dependence: The variation of $pH$ with temperature is generally small and often ignored.
Logarithmic nature: Since the $pH$ scale is logarithmic,a change of $1$ unit in $pH$ corresponds to a $10$-fold change in $[H^{+}]$.
Example: If $[H^{+}] = 1 \times 10^{-2} \ M$,then $pH = 2$. If $[H^{+}] = 1 \times 10^{-3} \ M$,then $pH = 3$.

The measurement of $pH$ of a solution is essential in various fields,including biological and cosmetic applications.
$1$. Rough Estimation using $pH$ Paper: The $pH$ of a solution can be estimated roughly using $pH$ paper,which exhibits different colors in solutions of different $pH$ values. As shown in the reference image,the $pH$ in the range of $0-14$ can be determined with an accuracy of $\sim 0.5$ using $pH$ paper.
$2$. Accurate Measurement using $pH$ Meter: For greater accuracy,$pH$ meters are used. $A$ $pH$ meter is a device that measures the $pH$-dependent electrical potential of the test solution with a precision of up to $0.001$. $pH$ meters as small as a writing pen are now commonly available.

(N/A) The relationships are as follows:
$1$. Neutral solution: $[H_3O^+] = [OH^-] = 1 \times 10^{-7} \ M$ and $pH = 7.0$.
$2$. Acidic solution: $[H_3O^+] > [OH^-]$,$[H_3O^+] > 10^{-7} \ M$,$[OH^-] < 10^{-7} \ M$,$pH < 7.0$ and $pOH > 7.0$.
$3$. Basic solution: $[H_3O^+] < [OH^-]$,$[H_3O^+] < 10^{-7} \ M$,$[OH^-] > 10^{-7} \ M$,$pH > 7.0$ and $pOH < 7.0$.

(N/A) The ionic equilibrium of dissociation of water at $298 \ K$ is given by:
$H_{2}O_{(l)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O^{+}_{(aq)} + OH^{-}_{(aq)}$
The ionic product of water is $K_{w} = [H_{3}O^{+}][OH^{-}] = 1.0 \times 10^{-14}$.
Taking the negative logarithm on both sides:
$-\log K_{w} = -\log ([H_{3}O^{+}][OH^{-}]) = -\log (1.0 \times 10^{-14})$
Using the property $\log(ab) = \log a + \log b$:
$-\log K_{w} = -\log [H_{3}O^{+}] - \log [OH^{-}] = 14.0$
Since $pH = -\log [H_{3}O^{+}]$ and $pOH = -\log [OH^{-}]$,and $pK_{w} = -\log K_{w}$:
$pK_{w} = pH + pOH = 14.0$
Thus,the relation is $pH + pOH = 14.0$ at $298 \ K$.

(A) For an aqueous solution at $25^{\circ}C$,the relationship between $pH$ and $pOH$ is given by: $pH + pOH = 14$.
Given $pH = 10$,we have $pOH = 14 - 10 = 4$.
The concentration of hydroxide ions is calculated as: $[OH^{-}] = 10^{-pOH} = 10^{-4} \ M$.
Thus,the concentration is $1 \times 10^{-4} \ M$.

(N/A) The concentration of $[H^{+}]$ is calculated using the formula $[H^{+}] = 10^{-pH}$.
For $pH = 12.0$,$[H^{+}] = 10^{-12.0} = 1 \times 10^{-12} \ M$.
For $pH = 5.6$,$[H^{+}] = 10^{-5.6} = 10^{0.4} \times 10^{-6} \approx 2.51 \times 10^{-6} \ M$.

(N/A) Given: $[H_3O^{+}] = 3.5 \times 10^{-8} \ M$.
Using the ion product of water at $298 \ K$,$K_w = [H_3O^{+}][OH^{-}] = 1.0 \times 10^{-14}$.
$[OH^{-}] = \frac{K_w}{[H_3O^{+}]} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-8}} \approx 2.86 \times 10^{-7} \ M$.
$pH = -\log[H_3O^{+}] = -\log(3.5 \times 10^{-8}) = 8 - \log(3.5) \approx 8 - 0.544 = 7.456$.

(A) Given,$[H^{+}] = 0.001 \ M = 10^{-3} \ M$.
$pH = -\log[H^{+}] = -\log(10^{-3}) = 3.0$.
Using the ionic product of water,$K_w = [H^{+}][OH^{-}] = 10^{-14}$ at $298 \ K$.
$[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \ M$.

(A) The relationship between $pH$ and $[H^+]$ is given by the formula: $pH = -\log[H^+]$.
Substituting the given $pH$ value: $5.4 = -\log[H^+]$.
Therefore,$[H^+] = 10^{-5.4}$.
$[H^+] = 10^{0.6} \times 10^{-6}$.
Since $10^{0.6} \approx 3.98$,the concentration is $[H^+] = 3.98 \times 10^{-6} \ M$.

(A) Given that the concentration of ${H_3}{O^+}$ is $[H_3O^+] = 1.0 \times 10^{-10} \ M$.
First,calculate the $pH$ of the solution:
$pH = -\log[H_3O^+] = -\log(1.0 \times 10^{-10}) = 10.0$.
At $25^{\circ} C$,the relationship between $pH$ and $pOH$ is given by $pH + pOH = 14.0$.
Therefore,$pOH = 14.0 - pH = 14.0 - 10.0 = 4.0$.

(B) The auto-ionization of water is an endothermic process.
According to Le Chatelier's principle,an increase in temperature shifts the equilibrium in the forward direction.
Therefore,the ionic product of water $(K_w = [H^+][OH^-])$ increases as the temperature increases.
At $298 \ K$,$K_w = 1 \times 10^{-14}$.
Since $393 \ K > 298 \ K$,the value of $K_w$ at $393 \ K$ will be greater than $1 \times 10^{-14}$.

(D)
As we know,$pH = -\log [H^+]$.
Initial $pH = 2.0$,so initial $[H^+]_i = 10^{-2} \ M$.
Final $pH = 5.0$,so final $[H^+]_f = 10^{-5} \ M$.
The ratio of change is $\frac{[H^+]_f}{[H^+]_i} = \frac{10^{-5}}{10^{-2}} = 10^{-3}$.
Thus,the hydrogen ion concentration decreases by a factor of $1000$ (thousand-fold).

(B) The $pH$ of a solution is defined as $pH = -\log [H^{+}]$.
When the concentration $[H^{+}]$ changes by a factor of $1000$,the change in $pH$ is given by $\Delta pH = -\log(\frac{[H^{+}]_{final}}{[H^{+}]_{initial}})$.
Since the concentration changes by a factor of $1000$,we have $\frac{[H^{+}]_{final}}{[H^{+}]_{initial}} = 1000 = 10^3$.
Therefore,$\Delta pH = -\log(10^3) = -3$.
This means the $pH$ value decreases by $3$ units.

(A) The auto-ionization of water is an endothermic process: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
As temperature increases,the equilibrium constant $K_W$ increases according to Le Chatelier's principle.
Since $K_W = [H^+][OH^-]$ and at neutrality $[H^+] = [OH^-] = \sqrt{K_W}$,the concentration of $H^+$ ions increases.
Therefore,$pH = -\log[H^+]$ decreases as the temperature rises.

(B) The auto-ionisation of water is an endothermic process: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
According to Le Chatelier's principle,increasing the temperature shifts the equilibrium to the right,which increases the degree of ionisation of water.
Consequently,the concentration of $[H^+]$ ions increases,which leads to a decrease in the $pH$ value of pure water (since $pH = -\log[H^+]$).
Therefore,Statement-$I$ is incorrect and Statement-$II$ is correct.

(C) The ionic product of water is given by $K_w = [H^+][OH^-]$.
For pure water,$[H^+] = [OH^-]$,so $K_w = [H^+]^2$.
Given $K_w = 4 \times 10^{-14}$ at $50^{\circ} C$.
Therefore,$[H^+] = \sqrt{4 \times 10^{-14}} = 2 \times 10^{-7} \ M$.
The $pH$ is calculated as $pH = -\log[H^+] = -\log(2 \times 10^{-7}) = -(\log 2 + \log 10^{-7}) = -(0.301 - 7) = 6.699 \approx 6.7$.
Since the concentration of $H^+$ ions is equal to the concentration of $OH^-$ ions in pure water,the water remains neutral despite the $pH$ value being less than $7$.