Water hydrolysis and pH scale Questions in English

(D) The ionic product of water is defined as the product of the molar concentrations of $H^+$ and $OH^-$ ions in water.
$K_w = [H^+][OH^-]$
The unit of molar concentration is $mol \ L^{-1}$.
Therefore,the unit of $K_w = (mol \ L^{-1}) \times (mol \ L^{-1}) = mol^2 \ L^{-2}$.

(D) The auto-ionization of water is an endothermic process: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$; $\Delta H > 0$.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium to the right,thereby increasing the value of the ionic product of water $(K_w)$.

(B) For pure water,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}]$.
Given $[H_3O^{+}] = 10^{-6}\,M$,therefore $[OH^{-}] = 10^{-6}\,M$.
The ionic product of water is defined as $K_w = [H_3O^{+}][OH^{-}]$.
Substituting the values: $K_w = (10^{-6}) \times (10^{-6}) = 10^{-12}$.

(B) The auto-ionization of water is an endothermic process represented by the equation: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium to the right.
Therefore,the concentration of $H^+$ and $OH^-$ ions increases,which leads to an increase in the ionic product of water $(K_w)$.
Since $90\,^{\circ}C > 25\,^{\circ}C$,the value of $K_w$ at $90\,^{\circ}C$ must be greater than $10^{-14}$.
Among the given options,only $1 \times 10^{-12}$ is greater than $10^{-14}$.

(B) The ionic product of water is defined as $K_w = [H^+][OH^-]$.
Taking the negative logarithm on both sides,we get $-\log K_w = -\log [H^+] - \log [OH^-]$.
By definition,$pH = -\log [H^+]$ and $pOH = -\log [OH^-]$.
Therefore,$pH + pOH = pK_w$.

(D) The relationship between $pH$ and hydrogen ion concentration is given by the formula: $pH = -\log [H^{+}]$.
Given $pH = 5.4$,we have $5.4 = -\log [H^{+}]$,which implies $\log [H^{+}] = -5.4$.
To find $[H^{+}]$,we calculate the antilog: $[H^{+}] = 10^{-5.4}$.
$[H^{+}] = 10^{0.6} \times 10^{-6} \approx 3.98 \times 10^{-6} \ mol/L$.

(B) At $25\,^oC$,pure water is neutral,meaning the concentration of hydrogen ions is equal to the concentration of hydroxide ions: $[H^{+}] = [OH^{-}]$.
The ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.
Thus,$[H^{+}] \times [OH^{-}] = 10^{-14}$,which implies $[H^{+}]^2 = 10^{-14}$,so $[H^{+}] = 10^{-7} \, M$.
The $pH$ is calculated as $pH = -\log[H^{+}] = -\log(10^{-7}) = 7$.

(D) The $pH$ of the solution is given as $5$.
We know that $[H^{+}] = 10^{-pH} = 10^{-5} \ M$.
Using the relation $pH + pOH = 14$ at $25^{\circ}C$,we get $pOH = 14 - 5 = 9$.
The hydroxyl ion concentration is given by $[OH^{-}] = 10^{-pOH} = 10^{-9} \ M$.

(A) The formula for $pH$ is defined as $pH = - \log [H^{+}]$.
Given that $[H^{+}] = 0.01 \, M$,which can be written as $10^{-2} \, M$.
Substituting the value into the formula: $pH = - \log [10^{-2}]$.
Using the logarithmic property $\log(a^b) = b \log(a)$,we get $pH = -(-2) \log(10)$.
Since $\log(10) = 1$,the $pH = 2$.

(C) The $pH$ scale ranges from $0$ to $14$.
At $pH = 7$,the solution is neutral.
At $pH < 7$,the solution is acidic.
At $pH > 7$,the solution is basic.
Since the given $pH$ is $9.5$,which is greater than $7$,the solution is basic.

(B) The ionic product of water $(K_w)$ is defined as the product of the concentrations of hydrogen ions and hydroxide ions in water at a given temperature.
$K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$ at $298 \ K$.
Taking the negative logarithm on both sides:
$-\log K_w = -\log ([H^+][OH^-]) = -\log (10^{-14})$.
$-\log K_w = -\log [H^+] - \log [OH^-] = 14$.
Since $pH = -\log [H^+]$ and $pOH = -\log [OH^-]$,we get:
$pH + pOH = 14$.

(A) For an aqueous acid $A$,the concentration of $H^+$ ions decreases upon dilution,which causes the $pH$ to increase towards $7$.
For an aqueous base $B$,the concentration of $OH^-$ ions decreases upon dilution,which causes the $pOH$ to increase,and consequently,the $pH$ decreases towards $7$.

(C) At $37^{\circ}C$,the ionic product of water $(K_w)$ is approximately $2.4 \times 10^{-14}$.
$pH = -\log[H^+]$.
For a neutral solution,$[H^+] = \sqrt{K_w} = \sqrt{2.4 \times 10^{-14}} \approx 1.55 \times 10^{-7}$.
$pH = -\log(1.55 \times 10^{-7}) \approx 6.8$.
Therefore,the $pH$ of a neutral solution at human body temperature is $6.8$.

(A) Initial concentration of $[H^{+}] = 10^{-5} \ M$.
After diluting $100$ times,the concentration of $[H^{+}]$ becomes $10^{-5} / 100 = 10^{-7} \ M$.
Since the concentration is very low,the $[H^{+}]$ contribution from water $(10^{-7} \ M)$ cannot be neglected.
Total $[H^{+}] = 10^{-7} + 10^{-7} = 2 \times 10^{-7} \ M$.
$pH = -\log(2 \times 10^{-7}) = 7 - \log(2) = 7 - 0.3010 = 6.699$.
Since the $pH$ is slightly less than $7$,the solution remains slightly acidic,but it is often considered nearly neutral in many contexts. However,among the given options,it is closest to neutral.

(A) The formula for $pH$ is given by: $pH = -\log [H^{+}]$.
Given $pH = 7.4$,we have $7.4 = -\log [H^{+}]$,which implies $\log [H^{+}] = -7.4$.
To find $[H^{+}]$,we calculate the antilog: $[H^{+}] = 10^{-7.4}$.
$[H^{+}] = 10^{0.6} \times 10^{-8}$.
Since $10^{0.6} \approx 3.98 \approx 4$,the concentration is $[H^{+}] \approx 4 \times 10^{-8} \ M$.

(B) The relationship between $pH$ and hydrogen ion concentration is given by the formula:
$pH = -\log[H^{+}]$
Given that $pH = 4$,we have:
$4 = -\log[H^{+}]$
$\log[H^{+}] = -4$
$[H^{+}] = 10^{-4} \ mol/L$
Therefore,the correct option is $B$.

(A) Given that $[OH^{-}] = 10^{-7} \ M$.
$pOH = -\log[OH^{-}] = -\log(10^{-7}) = 7$.
We know that at $25^{\circ}C$,$pH + pOH = 14$.
Therefore,$pH = 14 - pOH = 14 - 7 = 7$.

(B) The $pH$ of a solution is defined as the negative logarithm of the hydrogen ion concentration in $mol \ L^{-1}$.
Mathematically,it is expressed as $pH = -\log_{10}[H^{+}]$.
Therefore,the correct option is $B$.

(C) The relationship between $pH$ and $pOH$ at $25^{\circ}C$ is given by the equation: $pH + pOH = 14$.
Given that $pOH = 6.0$.
Substituting the value: $pH = 14 - 6.0 = 8.0$.

(D) The $pH$ of a solution is defined as the negative logarithm of the hydrogen ion concentration in $mol \ L^{-1}$.
Mathematically,it is expressed as $pH = - \log [H^{+}]$.
Alternatively,since $\log \frac{1}{[H^{+}]} = - \log [H^{+}]$,the expression $pH = \log \frac{1}{[H^{+}]}$ is also equivalent. However,the standard $IUPAC$ definition is $pH = - \log [H^{+}]$.

(D) The $pH$ of a solution is defined as $pH = -\log[H^+]$.
Adding pure water (options $A$ and $B$) to $HCl$ increases the volume,which decreases the concentration of $H^+$ ions,thereby increasing the $pH$.
Adding more $HCl$ (option $C$) increases the concentration of $H^+$ ions,which decreases the $pH$.
Adding the same dilute $HCl$ (option $D$) does not change the concentration of $H^+$ ions in the original solution,as the ratio of moles to volume remains constant. Therefore,the $pH$ remains unchanged.

(C) In distilled water,the auto-ionization reaction is $2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq).$
Since the water is neutral,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}] = 1 \times 10^{-6} \, mol/L.$
The ionic product of water,$K_w,$ is defined as $K_w = [H_3O^{+}][OH^{-}].$
Substituting the given values: $K_w = (1 \times 10^{-6}) \times (1 \times 10^{-6}) = 1 \times 10^{-12}.$
Therefore,the correct option is $C.$

(D) Given that the $pOH$ of the solution is $10.0$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Therefore,$pH = 14 - 10.0 = 4.0$.
The hydrogen ion concentration $[H^{+}]$ is calculated using the formula $[H^{+}] = 10^{-pH}$.
Thus,$[H^{+}] = 10^{-4} \ mol \ L^{-1}$.

(C) The $pH$ of a solution is calculated using the formula: $pH = -\log[H^{+}]$.
Given $[H^{+}] = 10 \times 10^{-4} \, \text{moles/litre} = 10^{-3} \, \text{M}$.
Substituting the value into the formula: $pH = -\log(10^{-3})$.
$pH = -(-3) \times \log(10) = 3 \times 1 = 3$.

(B) The relationship between $pH$ and hydrogen ion concentration is given by the formula: $pH = - \log [H^+]$.
As the $pH$ value decreases,the term $-\log [H^+]$ decreases,which implies that the value of $[H^+]$ must increase.
Therefore,the hydrogen ion concentration increases.

(D) At $25\,^{\circ}C,$ the relationship between $pH$ and $pOH$ is given by the equation: $pH + pOH = pK_w = 14.0$.
Given that $pH = 4.0$.
Substituting the value into the equation: $4.0 + pOH = 14.0$.
Therefore,$pOH = 14.0 - 4.0 = 10.0$.

(B) The $pH$ of a solution is defined as $pH = -\log[H^{+}]$.
Given $pH = 0$,we have $0 = -\log[H^{+}]$,which implies $[H^{+}] = 10^0 = 1 \ M$.
$A$ solution with a high concentration of $H^{+}$ ions is strongly acidic.

(C) The concentration of $H^{+}$ ions from the acid is $10^{-10} \ M$.
Since the concentration is very low,the contribution of $H^{+}$ ions from the auto-ionization of water $(10^{-7} \ M)$ cannot be neglected.
The total $[H^{+}] = [H^{+}]_{\text{acid}} + [H^{+}]_{\text{water}} = 10^{-10} + 10^{-7} \approx 10^{-7} \ M$.
Therefore,the $pH = -\log[H^{+}] = -\log(10^{-7}) = 7$.
However,since the solution is acidic,the $pH$ must be slightly less than $7$.
Thus,the $pH$ lies between $6$ and $7$.

(D) In pure water,the auto-ionization reaction is: $2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)$.
At $298 \ K$,the ionic product of water is $K_w = [H_3O^{+}][OH^{-}] = 1 \times 10^{-14}$.
Since $[H_3O^{+}] = [OH^{-}]$ in pure water,let $[H_3O^{+}] = x$.
Therefore,$x^2 = 1 \times 10^{-14}$,which gives $x = 1 \times 10^{-7} \ g \ ion/litre$.
Thus,the concentration of hydronium ions is $1 \times 10^{-7} \ g \ ion/litre$.

(B) $pH + pOH = pK_w = 14$.
If the $pH$ of a solution is less than $7$,the solution is acidic.
In an acidic solution,the concentration of hydrogen ions $[H^+]$ is greater than the concentration of hydroxide ions $[OH^-]$.
If the $pH$ is equal to $7$,the solution is neutral.
If the $pH$ is greater than $7$,the solution is basic.

(B) The relationship between $pH$ and hydrogen ion concentration is given by the formula: $pH = -\log[H^{+}]$.
Given that $pH = 2$,we have $2 = -\log[H^{+}]$,which implies $\log[H^{+}] = -2$.
Taking the antilog on both sides,we get $[H^{+}] = 10^{-2} \ M$ or $1 \times 10^{-2} \ M$.

(C) The $pH$ of a solution is defined as $pH = -\log[{H^{+}}]$.
For $pH = 3$,the concentration of ${H^{+}}$ ions is $[{H^{+}}]_{1} = 10^{-3} \ M$.
For $pH = 6$,the concentration of ${H^{+}}$ ions is $[{H^{+}}]_{2} = 10^{-6} \ M$.
The ratio of the concentrations is $\frac{[{H^{+}}]_{1}}{[{H^{+}}]_{2}} = \frac{10^{-3}}{10^{-6}} = 10^{3} = 1000$.
Therefore,the ${H^{+}}$ ion concentration is reduced by $1000$ times.

(B) $CO_2$ is an acidic oxide. When it dissolves in water,it forms carbonic acid $(H_2CO_3)$,which dissociates to release $H^+$ ions,thereby increasing the acidity and decreasing the $pH$ of the solution below $7$.

(D) The $pH$ of a solution is defined as $pH = -\log[H^{+}]$,which implies $[H^{+}] = 10^{-pH}$.
For the original solution with $pH = 2$,the concentration is $[H^{+}]_{1} = 10^{-2} \ M$.
For the final solution with $pH = 4$,the concentration is $[H^{+}]_{2} = 10^{-4} \ M$.
The ratio of the concentrations is $\frac{[H^{+}]_{1}}{[H^{+}]_{2}} = \frac{10^{-2}}{10^{-4}} = 10^{2} = 100$.
Therefore,the $[H^{+}]$ concentration must be decreased by a factor of $100$ to raise the $pH$ from $2$ to $4$.

(A) The $pH$ of a solution is defined as $pH = -\log[H^+]$.
For a $KOH$ solution,the ionic product of water $(K_w = [H^+][OH^-])$ increases with an increase in temperature.
As $K_w$ increases,the concentration of $[H^+]$ increases,which leads to a decrease in the $pH$ value.
Alternatively,$pH + pOH = pK_w$. As temperature increases,$pK_w$ decreases,and since $pOH$ is relatively stable for a strong base,$pH$ must decrease.

(C) The formula for $pH$ is given by $pH = -\log [H^{+}]$.
Given $[H^{+}] = 6 \times 10^{-4} \ M$.
$pH = -\log (6 \times 10^{-4}) = -(\log 6 + \log 10^{-4})$.
$pH = -(\log 2 + \log 3 - 4) = -(0.301 + 0.477 - 4) = -(-3.222) = 3.222$.
Rounding to two decimal places,the $pH$ is $3.22$.

(B) $pH = -\log [H^{+}]$
Given $[H^{+}] = 1.0 \times 10^{-6} \ mol/L$
$pH = -\log (1.0 \times 10^{-6}) = 6$
Therefore,the $pH$ value is $6$.

(A) The definition of $pH$ is given by the formula: $pH = -\log_{10}[H^{+}]$.
In this expression,the concentration of hydrogen ions,denoted as $[H^{+}]$,must be expressed in units of moles per litre ($mol \ L^{-1}$ or $M$).

(B) The $pH$ scale at $25 \, ^\circ C$ is defined as follows:
$pH < 7$ indicates an acidic solution.
$pH = 7$ indicates a neutral solution.
$pH > 7$ indicates a basic solution.
Since the given $pH$ is $6$,which is less than $7$,the solution is acidic.

(B) In pure water,the dissociation equilibrium is $H_2O \rightleftharpoons H^+ + OH^-$.
Since the concentration of $H^+$ ions equals the concentration of $OH^-$ ions,let $[H^+] = [OH^-] = x$.
The ionic product of water $(K_w)$ at $298 \ K$ is $1 \times 10^{-14}$.
Therefore,$K_w = [H^+][OH^-] = x^2 = 1 \times 10^{-14}$.
Solving for $x$,we get $x = \sqrt{1 \times 10^{-14}} = 1 \times 10^{-7} \ M$.

(A) Pure water $(H_2O)$ is a neutral substance at $25 \ ^{\circ}C$.
The concentration of $H^{+}$ ions in pure water is $10^{-7} \ M$.
Therefore,the $pH$ is calculated as: $pH = -\log[H^{+}] = -\log(10^{-7}) = 7$.
Thus,the correct option is $A$.

(C) The $pH$ of a solution is defined as $pH = -\log[H^{+}]$.
Let the initial concentration be $[H^{+}]_1 = C$,then $pH_1 = -\log(C)$.
If the concentration is increased by $10$ times,the new concentration is $[H^{+}]_2 = 10C$.
The new $pH$ is $pH_2 = -\log(10C) = -(\log(10) + \log(C)) = -(1 + \log(C)) = -1 - \log(C)$.
Comparing the two,$pH_2 = pH_1 - 1$.
Therefore,the $pH$ decreases by $1$.

(A) The relationship between $pH$ and hydrogen ion concentration is given by the formula: $[H^{+}] = 10^{-pH}$.
Given $pH = 4.58$,we have $[H^{+}] = 10^{-4.58}$.
To calculate this,we can write $10^{-4.58} = 10^{0.42 - 5} = 10^{0.42} \times 10^{-5}$.
Since $10^{0.42} \approx 2.63$,the concentration is $[H^{+}] = 2.63 \times 10^{-5} \, mol \, L^{-1}$.

(B) Given,$pH = 4$.
We know that at $25^{\circ}C$,$pH + pOH = 14$.
Therefore,$pOH = 14 - pH = 14 - 4 = 10$.
The concentration of hydroxide ions is given by $[OH^-] = 10^{-pOH}$.
Thus,$[OH^-] = 10^{-10} \ M$.

(B) The concept of $pH$ was first introduced by the Danish chemist,Soren Peder Lauritz Sorensen,at the Carlsberg Laboratory in $1909$.
It was later revised to the modern $pH$ definition in $1924$ to accommodate measurements in terms of electrochemical cells.

(A) The correct answer is both $(A)$ and $(B)$. At $pH = 7$,the concentration of $[H^+]$ and $[OH^-]$ ions are equal,which is characteristic of a neutral solution such as pure water at $25^{\circ}C$.

(B) Given,$[OH^{-}] = 1 \times 10^{-5} \ M$.
First,calculate the $pOH$ of the solution using the formula:
$pOH = - \log [OH^{-}] = - \log (1 \times 10^{-5}) = 5$.
At $25^{\circ}C$,the relationship between $pH$ and $pOH$ is given by:
$pH + pOH = 14$.
Substituting the value of $pOH$:
$pH = 14 - 5 = 9$.
Therefore,the correct option is $(B)$.

(C) Given: $pH = 3.82$.
We know that $pH = -\log [H^+]$.
Therefore,$[H^+] = 10^{-pH} = 10^{-3.82}$.
$[H^+] = 10^{0.18} \times 10^{-4}$.
Since $10^{0.18} \approx 1.51$,the concentration is approximately $1.5 \times 10^{-4} \ mol/L$.