Conduct a practical investigation to measure the pH of a range of acids and bases

[cs_content][cs_section parallax=”false” separator_top_type=”none” separator_top_height=”50px” separator_top_angle_point=”50″ separator_bottom_type=”none” separator_bottom_height=”50px” separator_bottom_angle_point=”50″ style=”margin: 0px;padding: 45px 0px;”][cs_row inner_container=”true” marginless_columns=”false” style=”margin: 0px auto;padding: 0px;”][cs_column fade=”false” fade_animation=”in” fade_animation_offset=”45px” fade_duration=”750″ type=”1/1″ style=”padding: 0px;”][cs_text]Acid/Base Reactions > Using Brønsted–Lowry Theory >[/cs_text][cs_text]Problem Statement: – Calculate the pH of the following solutions:

  1. 01M HCl
  2. 05M H2SO4
  3. 001M KOH

Solution:

1. pH of 0.01M HCl solution

Since HCl is a strong acid, it completely dissociates in aqueous solution[/cs_text][x_image type=”none” src=”https://easychem.com.au/wp-content/uploads/2019/06/pH-of-0.01M-HCl-solution.jpg” alt=”” link=”false” href=”#” title=”” target=”” info=”none” info_place=”top” info_trigger=”hover” info_content=””][/cs_column][/cs_row][/cs_section][cs_section parallax=”false” separator_top_type=”none” separator_top_height=”50px” separator_top_angle_point=”50″ separator_bottom_type=”none” separator_bottom_height=”50px” separator_bottom_angle_point=”50″ style=”margin: 0px;padding: 45px 0px;”][cs_row inner_container=”true” marginless_columns=”false” style=”margin: 0px auto;padding: 0px;”][cs_column fade=”false” fade_animation=”in” fade_animation_offset=”45px” fade_duration=”750″ type=”1/1″ style=”padding: 0px;”][cs_text]Therefore, the concentration of H+, [H+] will be 0.01M

Now, pH = – log[H+]

= – log (0.01)
= 2

2. pH of 0.05M H2SO4 solution

Since H2SO4 is a strong acid, it completely dissociates in aqueous solution[/cs_text][x_image type=”none” src=”https://easychem.com.au/wp-content/uploads/2019/06/pH-of-0.05M-H2SO4-solution.jpg” alt=”” link=”false” href=”#” title=”” target=”” info=”none” info_place=”top” info_trigger=”hover” info_content=””][/cs_column][/cs_row][cs_row inner_container=”true” marginless_columns=”false” style=”margin: 0px auto;padding: 0px;”][cs_column fade=”false” fade_animation=”in” fade_animation_offset=”45px” fade_duration=”750″ type=”1/1″ style=”padding: 0px;”][cs_text]Therefore, [H+] = 0.01M

Now, pH = – log[H+]

= – log (0.1)
= 1

3. pH of 0.001M KOH solution

Since KOH is a strong base, it completely dissociates in aqueous solution[/cs_text][x_image type=”none” src=”https://easychem.com.au/wp-content/uploads/2019/06/pH-of-0.001M-KOH-solution.jpg” alt=”” link=”false” href=”#” title=”” target=”” info=”none” info_place=”top” info_trigger=”hover” info_content=””][/cs_column][/cs_row][cs_row inner_container=”true” marginless_columns=”false” style=”margin: 0px auto;padding: 0px;”][cs_column fade=”false” fade_animation=”in” fade_animation_offset=”45px” fade_duration=”750″ type=”1/1″ style=”padding: 0px;”][cs_text]Therefore, [OH–] = 0.01M

Now, pOH = – log[OH–]

= – log (0.001)
= 3

Hence, pH = 14 – pOH = 14 –  3 = 11[/cs_text][/cs_column][/cs_row][/cs_section][/cs_content]

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