MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

I think the easiest place to see the $22$ is in a Kummer surface. Let $A$ be an abelian surface, so topologically $(S^1)^4$. This clearly has $h_2 = \binom{4}{2} = 6$, and there are obvious topological repreentatives for the $2$-cycles, given by $(S^1)^2$ in $6$ different ways.
Let $X$ be the quotient of $A$ by negation. This has $16$ singular points; the images of the $16$ $2$-torsion points of $A$. Let $Y$ be $X$ blown up at these $16$ points. Then $H_2(Y)$ is (ADDED rationally, see below) generated by the pushforwards of the $6$ $2$-cycles from $A$, and the $16$ $\mathbb{P}^1$'s introduced by resolving the singularities. $6+16=22$.
I think the easiest place to see the $22$ is in a Kummer surface. Let $A$ be an abelian surface, so topologically $(S^1)^4$. This clearly has $h_2 = \binom{4}{2} = 6$, and there are obvious topological repreentatives for the $2$-cycles, given by $(S^1)^2$ in $6$ different ways.
Let $X$ be the quotient of $A$ by negation. This has $16$ singular points; the images of the $16$ $2$-torsion points of $A$. Let $Y$ be $X$ blown up at these $16$ points. Then $H_2(Y)$ is generated by the pushforwards of the $6$ $2$-cycles from $A$, and the $16$ $\mathbb{P}^1$'s introduced by resolving the singularities. $6+16=22$.