Question

Calculate the pH of the following solutions:
$2.21g$ of $TlOH$ dissolved in water to give 2 litre of solution. (Assume $TlOH$ to be a strong base).
$0.49\% w/v$ ${H_2}S{O_4}$ solution.

Answer 1

We can calculate the pH of the solution using the concentration of hydroxide ions and concentration of hydrogen ion. The concentrations of ions are calculated from the values of the number of moles of the solution, volumes of the solution.
Formula used: The formula to calculate $pH$ of the solution is,
$pOH + pH = 14$
$pH = 14 - pOH$
From the concentration of hydrogen ions, the pH of the solution is calculated using the formula,
$pH = - \log \left[ {{H^ + }} \right]$

Complete step by step answer:
Given data contains,
Mass of $TlOH$ is $2.21g$
Volume of the solution is two liter.
First, let us calculate the moles of $TlOH$.
We know that $221.39g/mol$ is the molar mass of $TlOH$
The moles of $TlOH$ can be calculated using the grams and molar mass of $TlOH$.
Moles of $TlOH$ =$\dfrac{{Grams}}{\text{Molar mass}}$
Let us substitute the values of grams and molar mass of $TlOH$.
Moles of $TlOH$=$\dfrac{{Grams}}{\text{Molar Mass}}$
Substituting the known values we get,
Moles of $TlOH$=$\dfrac{{2.21g}}{{221.39g/mol}}$
On simplifying we get,
Moles of $TlOH$=$9.9 \times {10^{ - 3}}mol$
The moles of $TlOH$ is $9.9 \times {10^{ - 3}}mol$.
Let us now calculate the concentration using the moles of $TlOH$ and the volume of the solution.
Concentration$= \dfrac{{{\text{No}}{\text{. of moles}}}}{{{\text{Volume of solution}}}}$
Let us now substitute the values of moles of $TlOH$ and volume of the solution.
Concentration= $\dfrac{{9.9 \times {{10}^{ - 3}}}}{2}$
On simplifying we get,
Concentration= $4.95 \times {10^{ - 3}}M$
The concentration of $TlOH$ is $4.95 \times {10^{ - 3}}M$.
So, the concentration of $O{H^ - }$ ion is $4.95 \times {10^{ - 3}}M$.
Let us now calculate $pOH$ of the solution. The $pOH$ is calculated as,
$pOH = - \log \left[ {O{H^ - }} \right]$
Let us now substitute the value of concentration of $O{H^ - }$ ion to get $pOH$ of the solution.
$pOH = - \log \left[ {O{H^ - }} \right]$
$pOH = - \log \left[ {4.95 \times {{10}^{ - 3}}} \right]$
On simplifying we get,
$pOH = 2.3$
The $pOH$ of the solution is $2.3$.
Using the $pOH$ of the solution, we can calculate the $pH$ of the solution.
$pH = 14 - pOH$
Let us now substitute the value of $pOH$ to get the $pH$ of the solution.
$pH = 14 - pOH$
Substituting the value we get,
$\Rightarrow pH = 14 - 2.3$
On simplifying we get,
$\Rightarrow pH = 11.7$
The $pH$ of the solution is $11.7$.
Given data contains,
Weight/volume percent of ${H_2}S{O_4}$ solution is $0.49\%$.
We have to know that $0.49\% w/v$ means $0.49gm$ of ${H_2}S{O_4}$ is dissolved in $100mL$ ${H_2}S{O_4}$ .
Let us now calculate the concentration of ${H_2}S{O_4}$ using the molar mass and we should convert the volume in milliliters to liters by multiplying with 1000.
We know that $98g/mol$ is the molar mass of sulfuric acid.
Concentration of ${H_2}S{O_4}$=$\dfrac{{0.49 \times 1000}}{{98 \times 100}}$
Concentration of ${H_2}S{O_4}$=$0.05M$
The concentration of ${H_2}S{O_4}$ is $0.05M$.
We know that sulfuric acid dissociates into ${H^ + }$ and $S{O_4}^{2 - }$ . The dissociation equation is written as,
${H_2}S{O_4} \rightleftharpoons 2{H^ + } + S{O_4}^{2 - }$
We have to multiply the concentration by two as there are two hydrogen ions.
Concentration of ${H^ + }$=$2 \times M$
Concentration of ${H^ + }$=$2 \times 0.05$
Concentration of ${H^ + }$=$0.1M$
The concentration of ${H^ + }$ is $0.1M$.
We can calculate the $pH$ of the solution using the concentration of ${H^ + }$ .
$pH = - \log \left[ {{H^ + }} \right]$
Let us now substitute the value of concentration of ${H^ + }$ in the expression to get the $pH$ of the solution.
$pH = - \log \left[ {{H^ + }} \right]$
$pH = - \log \left[ {0.1} \right]$
$pH = 1$
The $pH$ of the solution is $1$.

Note:
We have to remember that for base, we have to calculate the $pH$ using the $pOH$ of the solution. For acid, we have to calculate the $pH$ using the concentration of ${H^ + }$ . So, while calculating the concentration which is nothing but the molarity, we have to calculate the moles (if grams are given) and volume. In case the volume is given milliliters, we have to convert milliliters to liters. In case if the conversion of milliliters to liters is not done, there would error in $pH$ .