Question

A)  0.50g of hydrogen chloride (HCl ) is dissolved in water to make 2.5L   of solution. What is the pH  of the resulting hydrochloric acid solution? Express the pH numerically to two decimal places.

B)  0.15g   of sodium hydroxide (NaOH ) pellets are dissolved in water to make 4.0L   of solution. What is the pH  of this solution?Express the pH numerically to two decimal places.

Answer 1

Concepts and reason

The concept used to solve this question is to calculate the pH of the given HCl solution and NaOH solutions. To calculate the pH of the HCl solution, the concentration of the solution is calculated. To calculate the pH of the NaOH solution, pOH of the NaOH solution is calculated from the concentration of the solution.

Fundamentals

Concentration of the solution can be calculated by the molarity equation which is defined as the number of moles of compound per liter of solution. This is also known as molar concentration.

${\rm{Molarity = }}\frac{{{\rm{Number of moles of solute}}}}{{{\rm{Volume (in liters)}}}}$

pH: The pH of the solution can be defined as the negative logarithm of concentration of ${{\rm{H}}^ + }$ ions in the solution.

${\rm{pH = - log[}}{{\rm{H}}^{\rm{ + }}}{\rm{]}}$

Correspondingly, the pOH can be given by the following formula.

${\rm{pOH = - log[O}}{{\rm{H}}^ - }{\rm{]}}$

The pH and pOH are related by the following expression

${\rm{pH = 14 - pOH}}$

A)

Given,

The weight of hydrogen chloride (HCl) dissolved = 0.5 g.

The volume of water taken = 2.5 L.

The molecular weight of hydrogen chloride (HCl) = ${\rm{1}}{\rm{.01 + 35}}{\rm{.45 = 36}}{\rm{.46 g/mol}}$

$\begin{array}{c}\\{\rm{The number of moles of HCl = }}\frac{{{\rm{Weight}}}}{{{\rm{Moelcular weight}}}}\\\\ = {\rm{ }}\frac{{{\rm{0}}{\rm{.5 g}}}}{{{\rm{36}}{\rm{.46 g/mol}}}}\\\\ = {\rm{ 0}}{\rm{.0137 moles}}\\\end{array}$

$\begin{array}{c}\\{\rm{The concentration of HCl solution = }}\frac{{{\rm{Number of moles of HCl}}}}{{{\rm{Total volume of water in liter}}}}\\\\ = \frac{{{\rm{0}}{\rm{.0137 moles}}}}{{{\rm{2}}{\rm{.5 L}}}}{\rm{ }}\\\\{\rm{ = 0}}{\rm{.00548 moles/L}}\\\\{\rm{ = 5}}{\rm{.4 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ M}}\\\end{array}$

The concentration of HCl solution = the concentration of ${{\rm{H}}^ + }$ ions.

By substituting the value of [ ${{\rm{H}}^ + }$ ], the pH of the given HCl solution can be calculated as follows:

$\begin{array}{c}\\{\rm{pH = - log[}}{{\rm{H}}^{\rm{ + }}}{\rm{]}}\\\\{\rm{ = - log(5}}{\rm{.4 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{)}}\\\\{\rm{ = 2}}{\rm{.26}}\\\end{array}$

Therefore, the pH of HCl solution is 2.26.

B)

Given,

The weight of sodium hydroxide (NaOH) pellets dissolved = 0.15 g.

The volume of water taken = 4.0 L.

The molecular weight of sodium hydroxide (NaOH) = ${\rm{40}}{\rm{.0 g/mol}}$

$\begin{array}{c}\\{\rm{The number moles of NaOH = }}\frac{{{\rm{Weight}}}}{{{\rm{Moelcular weight}}}}\\\\ = {\rm{ }}\frac{{{\rm{0}}{\rm{.15 g}}}}{{{\rm{40 g/mol}}}}\\\\ = {\rm{ 0}}{\rm{.00375 moles}}\\\end{array}$

$\begin{array}{c}\\{\rm{The concentration of NaOH solution = }}\frac{{{\rm{Number of moles of NaOH}}}}{{{\rm{Total volume of water in litre}}}}\\\\ = \frac{{{\rm{0}}{\rm{.00375 moles}}}}{{{\rm{4}}{\rm{.0 L}}}}{\rm{ }}\\\\{\rm{ = 0}}{\rm{.000937 moles/L}}\\\\{\rm{ = 9}}{\rm{.3 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ M}}\\\end{array}$

The concentration of NaOH solution = the concentration of ${\rm{O}}{{\rm{H}}^ - }$ ions.

By substituting the value of [ ${\rm{O}}{{\rm{H}}^ - }$ ], the pOH of the given NaOH solution can be calculated as follows:

$\begin{array}{c}\\{\rm{pOH = - log[O}}{{\rm{H}}^ - }{\rm{] }}\\\\{\rm{ = - log(9}}{\rm{.3 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{)}}\\\\{\rm{ = 3}}{\rm{.04}}\\\end{array}$

The pH of the NaOH solution is calculated as follows:

$\begin{array}{c}\\{\rm{pH = }}14 - {\rm{pOH}}\\\\{\rm{ = 14 - 3}}{\rm{.04}}\\\\{\rm{pH = 10}}{\rm{.96}}\\\end{array}$

Therefore, the pH of NaOH solution is 10.96.

Ans: Part A

Part A

Answer

The pH of the HCl solution is 2.26.

Part B

The pH of NaOH solution is 10.96.