You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.200 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer?

Answer : The volume of sodium benzoate and benzoic acid  solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

Explanation :

Let the volume of sodium benzoate (salt) be, x

So, the volume of benzoic acid  (acid) will be, (100 - x)

Using Henderson Hesselbach equation :

Now put all the given values in this expression, we get:

x = 29.0

The volume of sodium benzoate = x = 29.0 mL

The volume of benzoic acid  (acid) = (100 - x) = (100 - 29.0) = 71 mL

Thus, the volume of sodium benzoate and benzoic acid  solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer