$p{K}_a=6\mathrm{.}86$

$pH=7\mathrm{.}00$

$[H_{2}P{O}_4^{-}]=0\mathrm{.}19M$

$pH=p{K}_a+log(\frac{[A^{-}]}{[HA]})=7\mathrm{.}00$

$7\mathrm{.}00=6\mathrm{.}86+log(\frac{[A^{-}]}{0\mathrm{.}19})$

$0\mathrm{.}14=log(\frac{[A^{-}]}{0\mathrm{.}19})$

$[A^{-}]=0\mathrm{.}26M$

Part B

Now assume you wish to make a buffer at the same $pH,$ using the same substances, but want the total phosphate molarity $([HP{O}_4^{2-}]+[H_{2}P{O}_4^{-}])$to equal $0\mathrm{.}39M.$ What concentration of the $K{H}_{2}P{O}_4$ would be required?

Express your answer to two significant figures.

$\sqrt[]{}A$

$?$

Request Answer