The original question is to find the pH and the pOH of 0.023 M of perchloric acid.
Answer:
pH = 1.638
pOH = 12.362
Explanation:
1- getting the pH:
pH can be calculated using the following rule:
pH = -log[H+]
Since the given solution is an acid, this means that [H+] is the same as the concentration of the solution.
This means that:
[H+] = 0.023
Substitute in the above equation to get the pH as follows:
pH = -log[0.023]
pH = 1.638
2- getting the pOH:
We know that:
pH + pOH = 14
We have calculated that pH = 1.638.
Substitute in the above equation to get the pOH as follows:
pOH + 1.638 = 14
pOH = 14 - 1.638
pOH = 12.362
Hope this helps :)
Answer:
The answer is "0.00172172603".
Explanation:
Given:

Formula:


I had this same question on one of are worksheets is D I can’t really explain it but the answer is D
First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 =
![\frac{[H^+][HC2O4^-]}{[H2C2O4]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BHC2O4%5E-%5D%7D%7B%5BH2C2O4%5D%7D%20)
= x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6