The hydronium and hydroxide concentrations of a solution that is 5.0 x 10-3 M H2SO4 is 2.7.
pH= -log[H+] - (i)
10^-3=H2So4
H+= 2×10-3
here ,
h2so4 ——— 2[H+] + so4^2-
thus [H+]= 2*10^(-3) because hydrogen ion has two moles
pH= -log[H+]
pH= -log(2×10^-3)
pH= 3-log2
pH= 3-log2pH= 2.7
The pH is 2.7
<h3>What is pH?</h3>
PH is the degree of alkalinity and acidicity in a solution.
Therefore, The hydronium and hydroxide concentrations of a solution that is 5.0 x 10-3 M H2SO4 is 2.7
Learn more about pH from the link below.
https://brainly.in/question/9937410
Answer:
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
Explanation:
The SA node, the cardiac center in the medulla oblongata, and the endocrine system
