Answer :
Solution :
The relationship between pH and [tex]$[H_3O^+]$[/tex] is pH = [tex]$- \log [H_3O^+]$[/tex]
The concentration of [tex]$[H_3O^+]$[/tex] is given as
pH = [tex]$- \log [H_3O^+]$[/tex]
[tex]$[H_3O^+]=10^{-pH}$[/tex]
[tex]$=10^{-1.510}$[/tex]
[tex]$=3.09 \times 10^{-2}$[/tex]
Therefore, the concentration of [tex]$[H_3O^+]$[/tex] is [tex]$=3.09 \times 10^{-2}$[/tex]
The relationship between pH and [tex]$[H_3O^+]$[/tex] is pH = [tex]$- \log [H_3O^+]$[/tex]
The concentration of [tex]$[H_3O^+]$[/tex] is given as
pH = [tex]$- \log [H_3O^+]$[/tex]
[tex]$=- \log[0.15]$[/tex]
= 0.82
Therefore, the pH of the solution is 0.82
Now updating the ICE table for 0.15 M [tex]$CH_3COOH$[/tex] is as follows :
[tex]$CH_3COOH \ \ \ + \ \ \ H_2O \ \ \rightleftharpoons \ \ CH_3COO^- \ \ + \ \ H_3O^+$[/tex]
[tex]$I\ (M)$[/tex] 0.15 0 0
[tex]$C\ (M)$[/tex] -x +x +x
[tex]$E\ (M)$[/tex] 0.15 - x x x
∴ Acidionization constant,
[tex]$K_a=\frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$[/tex]
[tex]$1.82 \times 10^{-5}=\frac{(x)(x)}{(0.15-x)}$[/tex]
[tex]$1.82 \times 10^{-5}=\frac{x^2}{(0.15-x)}$[/tex]
We can assume that : (0.15 - x) M = 0.15 M
[tex]$1.82 \times 10^{-5}=\frac{x^2}{0.15}$[/tex]
[tex]$x^2=1.82 \times 10^{-5} \times 0.15$[/tex]
[tex]$x^2=2.7 \times 10^{-6}$[/tex]
[tex]$x=1.7 \times 10^{-3} \ M$[/tex]
Therefore, from the equilibrium table, [tex]$[H_3O^+] =x=1.7 \times 10^{-3} \ M$[/tex]
pH = [tex]$- \log [H_3O^+]$[/tex]
[tex]$=- \log(1.7 \times 10^{-3} \ M)$[/tex]
= 2.77
Hence the pH of 0.15 M [tex]$CH_3COOH$[/tex] is 2.77
Stronger acid always have a lower pH value.
The value of pH for the given solution are 0.0, 7.45, 0.82 and 2.77
So the increasing order of the acidity is :
pH = 0.0 > pH = 0.82 (0.15 M HCl) > pH = 2.77 (0.15 M [tex]$CH_3COOH$[/tex]) > pH =7.45
Hydronium ion concentration can be calculated to know the pH of the solution. The increasing order of acidity is 0, 0.82, 2.77, and 7.45.
What is pH?
pH is the negative logarithm of the hydronium ion concentration. The hydronium ion concentration can be calculated as:
[tex]\begin{aligned}\rm pH &= \rm - log [H_{3}O^{+}]\\\\&= 10^{-1.510}\\\\&= 3.09 \times 10^{-2}\end{aligned}[/tex]
Thus, the hydronium ion concentration is 0.0309.
The pH when the hydronium ion concentration is 0.15, can be calculated as,
[tex]\begin{aligned}\rm pH &= \rm - log [H_{3}O^{+}]\\\\&= \rm -log (0.15)\\\\&= 0.82\end{aligned}[/tex]
Thus, the pH of the solution is 0.82.
Now the acid ionization constant is calculated using the ICE table as,
[tex]\begin{aligned}\rm K_{a} &= \rm \dfrac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}\\\\1.82 \times 10^{-5} &= \rm \dfrac{(x)(x)}{(0.15-x)}\end{aligned}[/tex]
Assuming that (0.15- x) = 0.15 M then,
[tex]\begin{aligned} 1.82 \times 10^{-5} &= \dfrac{x^{2}}{(0.15-x)}\\\\\rm x^{2} &= 2.7 \times 10^{-6}\\\\&= 1.7 \times 10^{-3}\;\rm M\end{aligned}[/tex]
Now, calculating pH of 0.15 M acetic acid is calculated as:
[tex]\begin{aligned}\rm pH &= \rm - log [H_{3}O^{+}]\\\\&= \rm -log (1.7 \times 10^{-3})\\\\&= 2.77\end{aligned}[/tex]
The stronger the acid lower will be the pH value. The pH of the solutions is 0.0, 7.45, 0.82, and 2.77.
Therefore, the correct order of the increasing acidity is 0.0 > 0.82 > 2.77 > 7.45.
Learn more about pH here:
https://brainly.com/question/24586675