you have to prepare a ph 5.03 buffer, and you have the following 0.10m solutions available: hcooh, hcoona, ch3cooh, ch3coona, hcn, and nacn. How many milliliters of each solution would you use to make approximately a liter of the buffer?

Solution :

The buffer is the one that contains weak acid (the is nearly equal to the required pH) and the salt of its conjugate base.

The  values of the given weak acids are as follows :

HCOOH :  = 3.744

HCN : = 9.21

Since the required pH is 5.03, the suitable buffer is the mixture of the acetic acid and the salt of its conjugate base.

Let suppose the volumes of and are x and y mL respectively.

So the total volume of the buffer is 1000 mL.

∴   x+ y = 1000  ................(1)

Writing the Henderson-Hasselbalch equation for the given buffer solution :

  .............(2)

y = 1.9 x

Substituting the values of y in equation (1), we get

x + 1.9 x = 1000

x = 345

Putting the value of x in (1), we get

345 + y = 1000

y = 655

Therefore the volume of is 345 mL and the volume of is 655 mL.

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.