In the balanced reaction and using the mole ratio:
1 answer:
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Answer:
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 (<em>H-H</em>) 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 <em>H-H</em> 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:
- The formula of sodium glyoxylate monohydrate is C₂HO₃Na·H₂O, thus its molecular weight is 114 g/mol
- The formula of glyoxylic acid monohydrate is C₂H₂O₃·H₂O, thus its molecular weight is 92 g/mol
- Mass of glyoxylic acid:
2.50 L * 0.377 M * 92 g/mol = 86.71 g glyoxylic acid monohydrate.
- Mass of sodium glyoxylate:
2.50 L * 1.223 M * 114 g/mol = 348.56 g sodium glyoxylate monohydrate.
<em>Why does the problem ask that we use the monohydrate forms? </em>Because that's the available reagent in the laboratory.