1) You need to prepare 1.000L of 0.50 M phosphate buffer, pH 6.21. Use the Hende
ID: 971789 • Letter: 1
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.
Explanation / Answer
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 pKa =6.64 the reason it s KH2PO4 have two disotiantion constans
-H2PO4 ------------------------ -2HPO4 pKa1 = ? like the relation
-2HPO4------------------------- -3PO4 pKa2 = 6.64
-H2PO4 ----------------------- -3PO4 pKaR= pKa1 + pKa2
pH = 6.21 pKa =7.21 -log([AOH]/[A]=? -log([AOH]/[A] = -Ph +PKa like the relation -log([AOH]/[A] = -Ph +PKa where 0.65= -6.21+PKaR PKaR = 6.86
pH = 6.21 pKa =6.64 -log([AOH]/[A]=? -log([AOH]/[A] = -Ph +PKa
-log([AOH]/[A] =-6.21 + 6.86 = 0.65 =-log([AOH]/[A] = 0.65 = -log([AOH]/[A] replace A like At-x and [AOH] like x
x/At-x =0.65---- 0.65At-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
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.