Question

A student created four solutions by adding 4.00mL, 8.00mL, 12.00mL, and 16.00mL of 0.666M NaOH to 12.0mL of 1.00M weak acid. Then the student diluted each of the four solutions to 100.00mL with deionized water. The pH of each solution was 4.615, 5.072, 5.488, and 6.114, respectively. For each of the four solutions, how many of moles of undissociated weak acid exist at equilibrium?

Answer 1

A solution having 10.66 M NaoH and 1 M weak acid (For eample aceteic acid) called a acidic buffer.

The pH of given solutions are 4.615, 5.072, 5.488, and 6.114 respectively.

from above data we will find the [H]⁺ concentration of respected solutions using following formula,

pH=-log [H]⁺

[H]⁺= anti pH

Therefore, the concentration of [H]⁺ given solution are 4.12 X 10⁴, 1.18 X 10⁵, 3.0 X 10⁵, and 1.3 X 10⁶, respectively.

The number of moles of weak acid remain at equlibrium = concentration of [H]⁺ at equlibrium in the solution

therefore the number of moles of weak acid remain in the solutions are 4.12 X 10⁴, 1.18 X 10⁵, 3.0 X 10⁵, and 1.3 X 10⁶, respectively.