[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, pOH and [H+] of 0.01M CH3COOH solution when the percentage of dissociation is 12.5%.
Solution:
Since CH3COOH is a weak acid, it dissociates partially in aqueous solution as follows[/cs_text][x_image type=”none” src=”https://easychem.com.au/wp-content/uploads/2019/06/percentage-of-dissociation-is-12.5.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]Given,
Percentage of dissociation = 12.5%
so, [H+] = 0.01 × 12.5%
= 0.01 × 0.125M
= 1.25 × 10-3 M
Now, pH = – log[H+]
= – log (1.25 × 10-3)
= 2.903
and pOH = 14 – pH
= 14 – 2.903
= 11.097
Problem Statement:
Calculate the pH, pOH and [OH–] of 0.05M NH4OH solution when the percentage of dissociation is 7.5%.
Solution:
Since NH4OH is a weak base, it dissociates partially in aqueous solution as follows[/cs_text][x_image type=”none” src=”https://easychem.com.au/wp-content/uploads/2019/06/percentage-of-dissociation-is-7.5.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]Given,
Percentage of dissociation = 7.5%
so, [OH–] = 0.05 × 7.5%
= 0.05 × 0.075M
= 3.75 × 10-3 M
Now, pOH = – log[OH–]
= – log (3.75 × 10-3)
= 2.426
and pH = 14 – pOH = 14 – 2.426
= 11.574[/cs_text][/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_column][/cs_row][/cs_section][/cs_content]