How many grams of glyoxylic acid and sodium glyoxylate are needed to prepare 2.50 l of a 1.60 m buffer at ph 3.85? the pka of glyoxylic acid is 3.34. note: use the monohydrate forms, hcoco2h h20 and hcoco2na h20 number 113.78 g hcoco, h h2o number incorrect. you may not have included the mass of the water g hcoco, na h2 molecule in your calculation of the molar mass of sodium glyoxylate.

86.71 g of glyoxylic acid monohydrate, and 348.56 g of sodium glyoxylate monohydrate.

Explanation:

In order to solve this problem we need to use the Henderson–Hasselbalch equation:

pH= pka +

In this case [A⁻] is the concentration of sodium glyoxylate, and [HA] is the concentration of glyoxylic acid.

Using the Henderson–Hasselbalch (H-H) equation, the given pH and pka we can determine a relationship between [A⁻] and [HA]:

3.85 = 3.34 +  

0.51 =

Also from the problem, we know that

[A⁻] + [HA] = 1.60 M

We rearrange that equation to

[A⁻] = 1.60 M - [HA]

And replace the value of  [A⁻] in the H-H equation and solve for [HA]:

We substract 0.377 M to 1.60 M in order to calculate [A⁻].

1.60 M - 0.377 M= 1.223 M = [A⁻]

Lastly we calculate the mass of each reagent, using the concentration, volume and molecular weights:

2.50 L * 0.377 M * 92 g/mol = 86.71 g glyoxylic acid monohydrate.

2.50 L * 1.223 M * 114 g/mol = 348.56 g sodium glyoxylate monohydrate.

Why does the problem ask that we use the monohydrate forms? Because that's the available reagent in the laboratory.

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.