# Why is ionic strength (mu) calculated differently for these 2 examples?

In the following examples; I don't understand why $\ce{Na_3PO_4}$ is a much more complicated process when calculating $\mu$ compared to $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$. Specifically, I don't understand how to determine which species to include in the $\mu$ equation. All of the examples I've come across just behave just using water and the original species at most. What is more confusing also, is why doesn't the $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$ include the concentrations and charges of their conjugates as the last example with $\ce{Na_3PO_4}$ does.Thanks for any help!!! Here is the work:

$0.1000M$ of each reagent to start.

$\ce{NaH_2PO_4}$

$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] << 0.1M$$ $$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-1)^2)$$

$$\mu = \frac{1}{2}\sum([0.1M](+1)^2 + [0.1M](-1)^2)$$

$$\mu = 0.1000M$$

$\ce{NaH_2PO_4}$

$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] << 0.1M$$ $$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-2)^2)$$

$$\mu = \frac{1}{2}\sum([0.2M](+1)^2 + [0.1M](-1)^2)$$

$$\mu = 0.3000M$$

$\ce{Na_3PO_4}$

Without using Quadratic (which is just to keep things simple): $$[OH^-]=0.044M$$ $$[PO_4^{3-}]=0.056M$$ $$[HPO_4^{2-}]=0.044M$$ $$[H^+]=\frac{K_w}{0.044M}=2.27*10^{-13}M$$ $$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [PO_4^{3-}](-3)^2+[HPO_4^{2-}](-2)^2+[H^+](+1^2))$$ $$\mu = \frac{1}{2}\sum([0.3M](+1)^2 + [0.056M](-3)^2+[0.044M](-2)^2+[2.27*10^{-13}M](+1^2))$$ $$\mu=0.5120M$$