Question

Calculate the ${\rm{pH}}$of the following solutions:
a)${\rm{2}}\,{\rm{g }}$ of ${\rm{TlOH}}$ dissolved in water to give ${\rm{2}}\,{\rm{litre}}$ of solution.
b) ${\rm{0}}{\rm{.3}}\,{\rm{g }}$ of ${\rm{Ca(OH}}{{\rm{)}}_2}$dissolved in water to give $500\,{\rm{ml}}$of solution.
c) ${\rm{0}}{\rm{.3}}\,{\rm{g }}$ of ${\rm{NaOH}}$dissolved in water to give ${\rm{200}}\,{\rm{ml}}$of solution.
d) $1\,{\rm{ml }}$ of ${\rm{13}}{\rm{.6}}\,{\rm{M}}\,{\rm{HCl}}$is diluted with water to give $1\,{\rm{litre}}$of solution.

Answer 1

As we know that, ${\rm{pH}}$measures the solution is either acidic or basic. There are different range of measuring ${\rm{pH}}$as- if the ${\rm{pH}}$value is below seven, then the solution will be acidic, if above then solution will be basic and if equal then neutral. If we know the concentration of acids or bases then we can find out the ${\rm{pH}}$of the solution.

Complete step by step answer:
The ${\rm{pH}}$of above questions can be calculated in steps as-
Step ${\rm{1}}$-First of all we will calculate the concentration by the formula.
${\rm{Concentration}} = \dfrac{{{\rm{number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{of}}\,{\rm{compound}}}}{{{\rm{volume(Litre)}}}}$
Step $2$- then we write the equation of compound in water
Step $3$ -the we apply the formula for base as
${\rm{pOH}} = - {\rm{log}}\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]$
${\rm{pH}} = {\rm{14}} - {\rm{pOH}}$
And for acids
${\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^ + }} \right]$
Let’s we find the answers:
a) First, let’s calculate the number of moles of ${\rm{TlOH}}$ as follows;
number of moles of TlOH = $\dfrac{2g}{{221g/mole}}$
and volume${\rm{(V)}} = {\rm{2}}\,{\rm{litre}}$
putting above value in Step ${\rm{1}}$we get,

{\rm{C}} = \dfrac{{{\rm{0}}{\rm{.009}}\,{\rm{mole}}}}{{{\rm{2}}\,{\rm{litre}}}}\\\ = {\rm{4}}{\rm{.5 \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}} \end{array}$$ Now we will write an equation which represents concentration of hydroxide- $$\begin{array}{l} \,\,\,{\rm{TlOH}}\,\,\,\,\,\,\, \to {\rm{T}}{{\rm{l}}^{{\rm{ + 1}}}}{\rm{(aq) + O}}{{\rm{H}}^{{\rm{ - 1}}}}{\rm{(aq)}}\\\ \,\,\,\,\,{\rm{4}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{4}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\,\, \end{array}$$ We get $$\begin{array}{c} {\rm{pOH}} = - {\rm{log}}\left[ {4.5 \times {{10}^{ - 3}}} \right]\\\ {\rm{pOH}} = 2.35\,\,\,\,\,\,\,\,\,\,{\rm{(log}}\,{\rm{4}}{\rm{.5}} = {\rm{0}}{\rm{.65)}} \end{array}$$ $$\begin{array}{c} {\rm{pH}} = {\rm{14}} - {\rm{2}}{\rm{.35}}\\\ = {\rm{11}}{\rm{.65}} \end{array}$$ b) Similarly, Number of mole fraction of $Ca{(OH)_2}$ = $\dfrac{0.3g}{{74g/mole}}$ and volume$${\rm{(V)}} = {\rm{500 ml}}$$ putting above value in Step $${\rm{1}}$$we get, $$\begin{array}{c} {\rm{C}} = \dfrac{{{\rm{0}}{\rm{.0041}}\,{\rm{mole}}}}{{0.5\,{\rm{litre}}}}\\\ = {\rm{8}}{\rm{.11 \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}} \end{array}$$ Now, we will write an equation which represents concentration of hydroxide- $$\begin{array}{l} \,\,\,{\rm{Ca(OH}}{{\rm{)}}_2}\,\,\,\,\,\,\, \to C{a^{{\rm{ + 2}}}}{\rm{(aq) + 2O}}{{\rm{H}}^{{\rm{ - 1}}}}{\rm{(aq)}}\\\ \,\,\,\,\,8.11{\rm{ \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \times 8.11{\rm{ \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\,\, \end{array}$$ We get, $$\begin{array}{l} {\rm{pOH}} = - {\rm{log}}\left[ {16.22 \times {{10}^{ - 3}}} \right]\\\ {\rm{pOH}} = 1.79\,\,\,\,\,\,\,\,\,{\rm{(log}}\,16.22{\rm{ = 1}}{\rm{.21)}} \end{array}$$ $$\begin{array}{c} {\rm{pH}} = {\rm{14}} - {\rm{1}}{\rm{.79}}\\\ = {\rm{12}}{\rm{.21}} \end{array}$$ c) For $${\rm{0}}{\rm{.3}}\,{\rm{g }}$$ of $${\rm{NaOH}}$$dissolved in water to give $${\rm{200}}\,{\rm{ml}}$$of solution. Number of moles NaOH =$\dfrac{0.3g}{{40g/mole}}$ and volume$${\rm{(V) = 0}}{\rm{.2}}\,{\rm{litre}}$$ putting above value in Step $${\rm{1}}$$we get, $$\begin{array}{c} {\rm{C}} = \dfrac{{{\rm{0}}{\rm{.0075}}\,{\rm{mole}}}}{{0.2\,{\rm{litre}}}}\\\ = {\rm{3}}{\rm{.75 \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}} \end{array}$$ Now, we will write an equation which represents concentration of hydroxide- $$\begin{array}{l} \,\,\,{\rm{NaOH}}\,\,\,\,\,\, \to {\rm{N}}{{\rm{a}}^{{\rm{ + 1}}}}{\rm{(aq) + O}}{{\rm{H}}^{{\rm{ - 1}}}}{\rm{(aq)}}\\\ \,\,\,\,\,3.75{\rm{ \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \times 3.75{\rm{ \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}}\,\, \end{array}$$ We get $$\begin{array}{c} {\rm{pOH}} = - {\rm{log}}\left[ {7.5 \times {{10}^{ - 2}}} \right]\\\ {\rm{pOH}} = 1.124 \end{array}$$ $$\begin{array}{c} {\rm{pH}} = {\rm{14}} - {\rm{1}}{\rm{.124}}\\\ = {\rm{12}}{\rm{.8}} \end{array}$$ d) Finally, for $$1\,{\rm{ml }}$$ of $${\rm{13}}{\rm{.6}}\,{\rm{M}}\,{\rm{HCl}}$$is diluted with water to give $$1\,{\rm{litre}}$$of solution. $$\begin{array}{c} {\rm{number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{of}}\,{\rm{HC}}\,{\rm{l}}\, = {\rm{13}}{\rm{.6}}\,{\rm{M \times }}\,{\rm{1}}\,{\rm{ml}}\\\ {\rm{ = 13}}{\rm{.6}}\,{\rm{ \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{mole}} \end{array}$$ $$\begin{array}{c} {\rm{C}} = \dfrac{{{\rm{13}}{\rm{.6}}\, \times {\rm{1}}{{\rm{0}}^{ - 3}}{\rm{mole}}}}{{1\,{\rm{litre}}}}\\\ {\rm{ = 13}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 3}}{\rm{M}} \end{array}$$ $$\begin{array}{l} \,\,\,{\rm{HCl}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\rm{H}}^{{\rm{ + 1}}}}{\rm{(aq) + C}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{(aq)}}\\\ 13.6{\rm{ \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,13.6{\rm{ \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\,\, \end{array}$$ $$\begin{array}{c} {\rm{pH}} = - {\rm{log}}\left[ {13.6\, \times \,{{10}^{ - 3}}} \right]\\\ = 1.87 \end{array}$$ **Note:** More is the stronger acid, smaller will be the $${\rm{pH}}$$ and more stronger the base, larger the value of $${\rm{pH}}$$.