Question

1) You need to prepare 1.000L of 0.50 M phosphate buffer, pH 6.21. Use the Henderson-Hasselbalch equation with a value of 6.64 for pK2, to calculate the quantities of K2PO4 and KH2PO4 you need to add to your flask.
What is the ratio you need of [K2HPO4]/[KH2PO4] .
For this question I know there is an answer already up, but it doesn't show work to help me understand how they got that answer. I also got something different using the HH equation.
6.21=6.64- log [K2HPO4]/[KH2PO4]
-0.43
10^(-0.43)= 0.372
but their answer is 0.65

How many moles of K2HPO4 and KH2PO4 will you need.
I got 0.36M and 0.14 M.
Using 0.372x=0.50-x
1.372x=0.50
x= 0.3644

but their answer is 0.30 and 0.20.
Please, please help me understand.

Answer 1

ok going to help you step by step to answer First going to do all possible

you have do a nice work but you never goint to get the answer whit this pK_a =6.64 the reason it s KH₂PO₄ have two disotiantion constans

^-H₂PO₄ ------------------------ ^-2HPO₄   pK_a1 = ? like the relation

^-2HPO₄------------------------- ^-3PO_{4 p}K_a2 = 6.64

^-H₂PO₄ -----------------------  ^-3PO_{4 p}K_aR= _pK_{a1 + p}K_a2

pH = 6.21 pK_a =7.21 -log([AOH]/[A]=?   -log([AOH]/[A] = -Ph +PK_a like the relation  -log([AOH]/[A] = -Ph +PK_a where 0.65= -6.21+PK_aR   PK_aR = 6.86

pH = 6.21 pK_a =6.64 -log([AOH]/[A]=?   -log([AOH]/[A] = -Ph +PK_a

-log([AOH]/[A] =-6.21 + 6.86 = 0.65 =-log([AOH]/[A] = 0.65 = -log([AOH]/[A] replace A like A_t-x and [AOH] like x

x/A_t-x =0.65---- 0.65A_t-0.65x = x =) X= 0.65*0.5mol/1.65X = 0.196mol of K2PO4 and 0.304 of KH2PO4

you have do a nice work