Question

You select a 150.0 ml volumetric flask to which you add 80.00 mL of a 2.30 M solution of the weak acid you just selected. To finish preparing your buffer you must now add the sodium salt of the conjugate base. How many grams of the sodium salt of your conjugate base must you add so that when you finally fill the flask to the mark with deionized water the buffer will have a pH of 8.50.

Answer 1

7.33 grams of NaOH is the amount of conjugate base added.

Step-by-step explanation:

given that:

total volume = 150 ml

Vacid = 80 ml

Macid = 2.30 M

Vbase= 150-80= 70ml

Mbase=?

atomic mass of NaOH = 39.99 grams/mole

formula for titration

Macid X Vacid = Mbase X Vbase

putting the values in the equation:

Mbase =
`(2.30 x 80)/(70)`

= 2.62M is the molarity of NaOH

Number of moles of

M =
`(number of moles)/(volume in litres)`

2.62 x 0.07 = number of moles

0.1834 moles of NaOH is there, to know the mass formula used is

mass = number of moles x atomic mass

= x 0.1834

= 7.33 grams of NaOH is the amount of conjugate base added.