Water hydrolysis and pH scale Questions in English

(C) Auto-protolysis (or self-ionization) of water involves the transfer of a proton from one water molecule to another.
The process can be represented as:
$H_2O(l) \rightleftharpoons H^{+}(aq) + OH^{-}(aq)$
$H_2O(l) + H^{+}(aq) \rightarrow H_3O^{+}(aq)$
Combining these,the net reaction is:
$2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)$
Thus,the products are hydronium ions $(H_3O^{+})$ and hydroxide ions $(OH^{-})$.

(C) Given $[H^{+}] = 0.05 \ M = 5 \times 10^{-2} \ M$.
We know that the ionic product of water at $25^\circ C$ is $K_w = [H^{+}][OH^{-}] = 10^{-14}$.
Therefore,$[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{10^{-14}}{5 \times 10^{-2}} = 0.2 \times 10^{-12} = 2.0 \times 10^{-13} \ M$.

(D) The ionic product of water is given by $K_{w} = [H^{+}][OH^{-}] = 10^{-14} \ M^{2}$ at $298 \ K$.
Given $[H^{+}] = 0.001 \ M = 10^{-3} \ M$.
Therefore,$[OH^{-}] = \frac{K_{w}}{[H^{+}]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \ M$.

(D) In an aqueous solution at $298 \ K$:
$[H^{+}] \times [OH^{-}] = K_w = 10^{-14} \ (mol \ dm^{-3})^2$
Given $[OH^{-}] = 1 \times 10^{-12} \ mol \ dm^{-3}$
$[H^{+}] = \frac{K_w}{[OH^{-}]}$
$[H^{+}] = \frac{10^{-14}}{1 \times 10^{-12}} = 10^{-2} \ mol \ dm^{-3}$
Therefore,$[H^{+}] = 0.01 \ mol \ dm^{-3}$.

(C) The concentration of hydrogen ions $[H^{+}]$ is given by the product of the degree of dissociation $(\alpha)$ and the molar concentration $(C)$.
$[H^{+}] = \alpha \cdot C = 10^{-3} \cdot 0.01 = 10^{-5} \ M$.
Now,calculate the $pH$ using the formula $pH = -\log [H^{+}]$.
$pH = -\log(10^{-5}) = 5$.
Finally,calculate the $pOH$ using the relation $pH + pOH = 14$.
$pOH = 14 - 5 = 9$.

(B) The relationship between $pH$ and hydrogen ion concentration $[H^+]$ is given by the formula: $pH = -\log[H^+]$.
Given $pH = 3.76$,we have $3.76 = -\log[H^+]$.
Therefore,$\log[H^+] = -3.76$.
To find $[H^+]$,we take the antilog of $-3.76$: $[H^+] = 10^{-3.76}$.
$[H^+] = 10^{0.24} \times 10^{-4}$.
Since $10^{0.24} \approx 1.738$,the concentration is $[H^+] = 1.738 \times 10^{-4} \ mol \ dm^{-3}$.

(D) The $pH$ of a solution is defined as $pH = -\log[H_3O^+]$.
For $pH = 4$,$[H_3O^+]_1 = 10^{-4} \ M$.
For $pH = 5$,$[H_3O^+]_2 = 10^{-5} \ M$.
The change in concentration is $\frac{[H_3O^+]_1}{[H_3O^+]_2} = \frac{10^{-4}}{10^{-5}} = 10$.
Thus,the concentration decreases by $10$ times.

(A) The relationship between $pH$ and the concentration of hydronium ions is given by the formula: $pH = -\log[H_3O^{+}]$.
Given $pH = 2.8$,we have $2.8 = -\log[H_3O^{+}]$,which implies $\log[H_3O^{+}] = -2.8$.
To find $[H_3O^{+}]$,we take the antilog of $-2.8$: $[H_3O^{+}] = 10^{-2.8}$.
$[H_3O^{+}] = 10^{0.2} \times 10^{-3}$.
Since $10^{0.2} \approx 1.585$,the concentration is $[H_3O^{+}] = 1.585 \times 10^{-3} \ mol \ dm^{-3}$.

(D) Given: $pOH = 4.94$.
We know that,$pOH = -\log[OH^{-}]$.
Therefore,$[OH^{-}] = 10^{-pOH}$.
$[OH^{-}] = 10^{-4.94}$.
To solve this,we can write $10^{-4.94}$ as $10^{0.06 - 5} = 10^{0.06} \times 10^{-5}$.
Since $\text{antilog}(0.06) \approx 1.148$,we get $[OH^{-}] = 1.148 \times 10^{-5} \ M$.
Thus,the correct option is $(d)$.

(B) The $pH$ is calculated using the formula: $pH = -\log_{10}[H^{+}]$.
Given $[H^{+}] = 2.2 \times 10^{-6} \ M$.
$pH = -\log_{10}(2.2 \times 10^{-6}) = -(\log_{10}(2.2) + \log_{10}(10^{-6}))$.
$pH = -(\log_{10}(2.2) - 6) = 6 - \log_{10}(2.2)$.
Since $\log_{10}(2.2) \approx 0.34$,
$pH = 6 - 0.34 = 5.66$.

(D) Given $pOH = 11$.
We know that $pH + pOH = 14$.
Therefore,$pH = 14 - 11 = 3$.
Since $pH = -\log[H^{+}]$,it follows that $[H^{+}] = 10^{-pH} = 10^{-3} \ M$.

(C) $pH = -\log [H^{+}]$
$pH = -\log (1.342 \times 10^{-3})$
$pH = -(\log 1.342 + \log 10^{-3})$
$pH = -(0.1277 - 3)$
$pH = 3 - 0.1277 = 2.87$

(A) The relationship between $pH$ and $pOH$ at $25^{\circ}C$ is given by: $pH + pOH = 14$.
Given $pOH = 9$,we can calculate $pH$ as: $pH = 14 - 9 = 5$.
The concentration of hydronium ions $[H_3O^{+}]$ is related to $pH$ by the formula: $[H_3O^{+}] = 10^{-pH}$.
Substituting the value of $pH$: $[H_3O^{+}] = 10^{-5} \ M$.

(D) The $pH$ is calculated using the formula: $pH = -\log [H^{+}]$
Given $[H^{+}] = 0.0063 \ M = 6.3 \times 10^{-3} \ M$
$pH = -\log (6.3 \times 10^{-3})$
$pH = -(\log 6.3 + \log 10^{-3})$
$pH = -(\log 6.3 - 3)$
$pH = 3 - \log 6.3$
Substituting the value of $\log 6.3 = 0.7993$:
$pH = 3 - 0.7993 = 2.2007$
Rounding to the nearest option,the $pH$ is $2.2$.

(D) The $pH$ of a solution is calculated using the formula: $pH = -\log [H^{+}]$.
Given $[H^{+}] = 3.981 \times 10^{-7} \ M$.
Substituting the value: $pH = -\log (3.981 \times 10^{-7})$.
Using the property $\log (a \times b) = \log a + \log b$: $pH = -(\log 3.981 + \log 10^{-7})$.
$pH = -(\log 3.981 - 7)$.
$pH = 7 - \log 3.981$.
Given $\log 3.981 = 0.6000$,so $pH = 7 - 0.6000 = 6.4$.

(D) $pH + pOH = 14$ at $298 \ K$.
$pOH = 14 - pH = 14 - 9.95 = 4.05$.
Since $pOH = -\log_{10} \left[OH^{-}\right]$,we have $\left[OH^{-}\right] = 10^{-pOH}$.
$\left[OH^{-}\right] = 10^{-4.05} = 10^{-5 + 0.95} = 10^{-5} \times 10^{0.95}$.
Using $10^{0.95} \approx 8.91$,we get $\left[OH^{-}\right] = 8.91 \times 10^{-5} \ M$.

(A) The formula for $pH$ is given by: $pH = -\log_{10}[H^{+}]$.
Therefore,the concentration of hydrogen ions is: $[H^{+}] = 10^{-pH}$.
Substituting the given value: $[H^{+}] = 10^{-3.6}$.
To solve this,we can write: $[H^{+}] = 10^{-4 + 0.4} = 10^{0.4} \times 10^{-4}$.
Since $10^{0.4} \approx 2.51$,the concentration is $[H^{+}] = 2.51 \times 10^{-4} \ M$.

(C) $pH = -\log [H^{+}]$
$pH = -\log (3.98 \times 10^{-6})$
$pH = -(\log 3.98 + \log 10^{-6})$
$pH = -(0.60 - 6)$
$pH = 5.40$

(C) Given: Initial $pH = 3$.
Since $pH = -\log[H^{+}]$,we have $[H^{+}]_1 = 10^{-pH} = 1 \times 10^{-3} \ M$.
Using the dilution formula $M_1 V_1 = M_2 V_2$:
$(1 \times 10^{-3} \ M) \times (20 \ mL) = M_2 \times (100 \ mL)$.
$M_2 = \frac{1 \times 10^{-3} \times 20}{100} = 2 \times 10^{-4} \ M$.
Therefore,the final $H^{+}$ ion concentration is $2 \times 10^{-4} \ M$.

(C) The $pH$ of a solution is determined by the concentration of $H^+$ ions. Adding a solution with the same $pH$ (i.e.,same concentration of $H^+$ ions) will not change the $pH$ of the original solution.
For $20 \ mL$ of $0.1 \ N \ HCl$,the concentration of $H^+$ is $0.1 \ M$,which corresponds to $pH = -\log(0.1) = 1$.
Option $(c)$ provides $500 \ mL$ of $HCl$ with $pH = 1$. Since the concentration of $H^+$ ions in this solution is the same as the original solution $(0.1 \ M)$,adding it will not change the $pH$ of the original $20 \ mL$ of $0.1 \ N \ HCl$ solution.

(A) When $KBr$ is dissolved in water,it dissociates into $K^{+}$ and $Br^{-}$ ions.
These ions interact with water molecules through ion-dipole interactions,a process known as hydration.

(B) The dissociation of $H_2O$ is an endothermic process.
With an increase in temperature,the equilibrium shifts in the forward direction according to Le Chatelier's principle.
This leads to an increase in the concentration of $H^+$ and $OH^-$ ions.
Therefore,the ionic product of water $(K_w = [H^+][OH^-])$ increases with an increase in temperature.

(D) The $pH$ range $1-7$ represents acidic solutions,while the $pH$ range $7-14$ represents basic solutions.
$I$. Black coffee: Acidic $(pH \approx 5)$
$II$. $0.2 \ M \ NaOH$: Basic $(pH > 7)$
$III$. Lemon juice: Acidic $(pH \approx 2)$
$IV$. Lime water $(Ca(OH)_2)$: Basic $(pH > 7)$
$V$. Human saliva: Acidic $(pH \approx 6.5)$
$VI$. Tomato juice: Acidic $(pH \approx 4)$
Acidic solutions $(pH \ 1-7)$: $I, III, V, VI$ (Total $4$)
Basic solutions $(pH \ 7-14)$: $II, IV$ (Total $2$)
Therefore,the number of solutions is $4$ and $2$ respectively.

(C) The $pH$ values of the given solutions are as follows:
$I$. Black coffee: $pH \approx 5$ (Acidic)
$II$. $0.2 \ M \ NaOH$: $pH \approx 13.3$ (Basic)
$III$. Lemon juice: $pH \approx 2$ (Acidic)
$IV$. Lime water: $pH \approx 12$ (Basic)
$V$. Human Saliva: $pH \approx 6.5$ (Acidic)
$VI$. Tomato juice: $pH \approx 4$ (Acidic)
Solutions with $pH < 7$ are $I$,$III$,$V$,and $VI$.
Therefore,the total number of acidic solutions is $4$.

(C) small change in temperature has little impact on the $pH$ of a solution as the concentration of the $H^{+}$ ions does not change significantly.
Thus,statement $I$ is correct.
$pH = -\log [H^{+}]$.
If the hydrogen ion concentration $[H^{+}]$ changes by a factor of $100$,let the initial concentration be $[H^{+}]_1$ and the final be $[H^{+}]_2 = 100 [H^{+}]_1$.
The change in $pH$ is $\Delta pH = pH_2 - pH_1 = -\log (100 [H^{+}]_1) - (-\log [H^{+}]_1) = -\log (100) - \log [H^{+}]_1 + \log [H^{+}]_1 = -2$.
Thus,the $pH$ changes by $2$ units,not $1$ unit.
Therefore,statement $II$ is incorrect.

(A) The $pH$ scale ranges from $0$ to $14$ for aqueous solutions at $25^{\circ}C$,where higher $pH$ values indicate higher basicity.
$(A)$ Saturated solution of $NaOH$ is a strong base with a very high concentration of $OH^-$ ions,resulting in a $pH \approx 15$.
$(B)$ $1 \ M \ HCl$ is a strong acid,resulting in a $pH \approx 0$.
$(C)$ Human saliva is slightly acidic to neutral,with a $pH \approx 6.4$.
$(D)$ Lemon juice is acidic,with a $pH \approx 2.2$.
Therefore,the saturated solution of $NaOH$ has the highest $pH$ value. Hence,option $(A)$ is correct.

(B) The $pH$ of a solution is defined as the negative logarithm of the hydrogen ion concentration: $pH = -\log[H^+]$.
Given $[H^+] = 4.7 \times 10^{-4} \ M$.
$pH = -\log(4.7 \times 10^{-4})$.
Using the property $\log(a \times b) = \log a + \log b$,we get:
$pH = -(\log 4.7 + \log 10^{-4})$.
$pH = -(\log 4.7 - 4)$.
$pH = 4 - \log 4.7$.
Since $\log 4.7 \approx 0.672$,
$pH = 4 - 0.672 = 3.328 \approx 3.33$.

(D) At $25^{\circ} C$,the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.
For pure water,$[H^{ }] = [OH^{-}] = \sqrt{K_w} = 10^{-7} \ M$,which gives $pH = 7$.
The auto-ionization of water is an endothermic process $(\Delta H > 0)$.
According to Le Chatelier's principle,an increase in temperature shifts the equilibrium to the right,increasing the value of $K_w$.
At $80^{\circ} C$,$K_w > 10^{-14}$,which implies $[H^{ }] > 10^{-7} \ M$.
Since $pH = -\log[H^{ }]$,an increase in $[H^{ }]$ results in a decrease in $pH$.
Therefore,at $80^{\circ} C$,the $pH$ of pure water is $< 7.0$.

(C) The auto-ionization of water,$H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}$,is an endothermic process $(\Delta H > 0)$.
According to Le Chatelier's principle,an increase in temperature shifts the equilibrium to the right,increasing the value of the ionic product $(K_w)$.
Conversely,a decrease in temperature shifts the equilibrium to the left,decreasing the value of $K_w$.
Since the temperature is decreased from $35^{\circ} C$ to $10^{\circ} C$,the value of $K_w$ must be less than $1.96 \times 10^{-14}$.
Among the given options,$2.95 \times 10^{-15}$ is the only value smaller than $1.96 \times 10^{-14}$.

(C) The $10$ millionth part is equivalent to $10^{-7}$.
$\therefore$ The volume of the part is $1.33 \ cm^{3} \times 10^{-7} = 1.33 \times 10^{-7} \ mL$.
For pure water at $25^{\circ} C$,$[H^{+}] = 10^{-7} \ mol/L$.
Since $1 \ L = 1000 \ mL$,$1 \ mL$ of water contains $10^{-7} / 1000 = 10^{-10} \ mol$ of $H^{+}$ ions.
Therefore,the number of moles of $H^{+}$ in $1.33 \times 10^{-7} \ mL$ is $1.33 \times 10^{-7} \times 10^{-10} = 1.33 \times 10^{-17} \ mol$.
Number of $H^{+}$ ions $= \text{moles} \times N_{A} = 1.33 \times 10^{-17} \times 6.022 \times 10^{23}$.
$= 8.009 \times 10^{6} \approx 8.01 \times 10^{6}$ or $8.01$ million.