How many equivalents are in 0.40 mole of #K^+#?

1 Answer
Jul 18, 2016

Answer:

#"0.40 Eq"#

Explanation:

In the context of ions, an equivalent can be thought of as the number of moles of monovalent ions needed to neutralize the charge of an ion of opposite charge.

In your example, you're dealing with a potassium cation, #"K"^(+)#. This ion carries a #1+# charge, so right from the start you know that you're looking for the number of #1-# ions, i.e. monovalent ions of opposite charge, needed to neutralize this #1+# charge.

Since a charge of #1+# can be neutralized by a charge of #1-#, it follows that every mole of potassium cations will produce #1# mole of equivalents in solution.

Therefore, you will have

#0.40 color(red)(cancel(color(black)("moles K"^(+)))) * "1 Eq"/(1color(red)(cancel(color(black)("mole K"^(+))))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.40 Eq")color(white)(a/a)|)))#

A quick way to remember how this works -- the number of equivalents of a given ion in solution is calculated using the number of moles and the valence of said ion.

#color(blue)(|bar(ul(color(white)(a/a)"no. of Eq" = "no. of moles" xx "valence"color(white)(a/a)|)))#

For example, one mole of a #color(red)(2+)# cation like #"Mg"^color(red)(2+)# would produce

#"no. of Eq" = "1 mole" xx color(red)(2) = "2 Eq"#

in solution. This is the same as saying that you need #color(red)(2)# moles of #1-# anions to neutralize the charge on one mole of #color(red)(2+)# cations.