Let's check the phone application "buffers" for it's ability to calculate how to prepare a phosphate buffer.

Suppose I want $1$ liter of a $0\mathrm{.}1M$ phosphate buffer at a pH of $6\mathrm{.}8.$

Phosphate has a pK$2$ of $7\mathrm{.}2$

MW of monosodium phosphate $($aka monobasic$)$ is $119\mathrm{.}98$

MW of disodium phosphate $($aka dibasic$)$ is $141\mathrm{.}96$

You already calculated the fraction of dibasic we need from the $H-H$ equation. In other words you solved

for A$-/$total$.$

That fraction times $0\mathrm{.}1$ moles is the amount of dibasic, multiply by the MW to get the weight of dibasic

we need. Answer will be in grams

Calculate to $3$ sig figures