You need to prepare 1.000 L (in a volumetric flask) of 0.50 M phosphate buffer,

Question

You need to prepare 1.000 L (in a volumetric flask) of 0.50 M phosphate buffer, pH 6.23. Use the Henderson Hasselbalch equation with a value of 6.64 for pK2 to calculate the quantities of K2HPO4 and KH2PO4 you need to add to the flask. Record the steps of these calculations and use them as a guide for the calculation you will have to make in the laboratory. (Calculate the intermediate values to at least one more significant figure than required). 1.) What is the [K2HPO4]/[KH2PO4] ratio you will need? 2.) What concentrations of [K2HPO4] and [KH2PO4] will you need to make the total concentration 0.50 M? 3.) How many moles of K2HPO4 and KH2PO4 will you need? 4.) What mass of K2HPO4 and KH2PO4 will you need?

Explanation / Answer

let x represent [K2HPO4] and y represent [KH2PO4]
We know:

ph= 6.23

pka= 6.64

ph=pka +log(x/y)

log(x/y) = 6.23-6.64

= -0.41

x/y =0.38904514499

(1) x/y=0.389--> x=0.389y
(2) x+y=0.50M
Plugging equation (1) into equation (2) we get:

(0.389y)+y=0.50M

1.537y=0.50M

y= 0.3599712023038157M = [KH2PO4] =0.356
Now plug back into equation (1)
x=(0.389)(0.356)=0.14M = [K2HPO4]

Because it is in a 1L flask we can find moles by:

moles KH2PO4= (0.356 moles/L)*(1L)= 0.356 moles

moles K2HPO4= (0.14 moles/L)*(1L)= 0.14 moles

To find mass needed, use molar mass:

Mass KH2PO4:

0.356*136.1= 48.4516g

Mass K2HPO4

=0.14*174

=24.388 gm

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