Question #01c42

1 Answer
Doc048
Jan 9, 2018

#Cations# = #H^+##ions# = #1.204xx10^24 H^+#

Explanation:

#"Given"color(white)(.) "mass"color(white)(.) "of"# #H_2SO_4# = #196color(white)(.)grams#
=> #"moles"color(white)(.) "of"color(white)(,) H_2SO_4# = #(196g)/(98g/"mole"# = #2.13color(white)(.)##color(white)(.)"moles"##H_2SO_4#

Assuming 100% ionization:
=>#1color(white)(.)"mole"color(white)(.) H_2SO_4# => #2color(white)(.)"moles" color(white)(.)ions H^+# + #1color(white)(.)"mole" color(white)(.)SO_4^(2-)color(white)(.)"ions"#

=>#2.13color(white)(.) "moles"color(white)(.) H_2SO_4# => #4.26"color(white)(.)#moles# color(white)(.)H^+ions#
=>#4.26##color(white)(.)##cancel("mole"H^+ion")# #xx##(6.023xx10^23##H^+ioncolor(white)(.)"particles"# / #1.00color(white)(.)cancel("mole"H^+ions)#)

=> #1.204xx10^24##H^+ions#

NOTE: The #H^+# ion is also the 'Hydronium Ion' #H_3O^+#

Related questions
Impact of this question
748 views around the world
You can reuse this answer
Creative Commons License