Help! 7th Grade Math!

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 gl

yaroslaw [1]

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 + $log\frac{[A^{-} ]}{[HA]}$

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 + $log\frac[{A^{-}] }{[HA]}$

0.51 = $log\frac{[A^{-} ]}{[HA]}$

$10^{0.51} =\frac{[A^{-}] }{[HA]} \\3.24=\frac{[A^{-} ]}{[HA]}$

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]:

$3.24=\frac{1.60M-[HA]}{[HA]}\\3.24*[HA]=1.60-[HA]\\4.24*[HA]=1.60\\0.377 M = [HA]$

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.

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

6 0

4 years ago

The flavor and odor of fruits comes primarily from the esters they contain. For example, a major component in the flavor of oran

Lilit [14]

Answer:

Explanation:

Concepts and reason

This problem is based on the concept of hydrolysis of esters.

An ester is hydrolyzed to a carboxylic acid and an alcohol when treated with aqueous acid or aqueous base. Under alkaline conditions, the carboxylic acid is obtained in the form as its salt.

Fundamentals

Alkaline hydrolysis of ester is done with strong base {\\rm{NaOH}}NaOHand {\\rm{O}}{{\\rm{H}}^ - }OH\u2212acts as nucleophilic reagent. This reaction is reversible, since carboxylate anion has tendency to react with an alcohol and gives back ester.

Step-by-step

Step 1 of 2

Attack of [{ m{O}}{{ m{H}}^ - }]on carbonyl take place as follows resulting formation of tetrahedral intermediate:

\u043e\u043d\u043e\u043d

Explanation | Hint for next step

The {\\rm{O}}{{\\rm{H}}^ - }OH\u2212nucleophile attacks on the electrophilic carbon of an ester {\\rm{C}} = {\\rm{O}}C=O and forms tetrahedral intermediate after breaking the \\pi\u03c0-bond.

Step 2 of 2

Hydrolysis of product formed from step 1 followed by reaction with {\\rm{NaOH}}NaOHis as follows:

The products of the reaction are:

Explanation | Common mistakes

On the reaction of octyl acetate with aqueous sodium hydroxide, the products octyl alcohol and acetate ion are formed after omitting the {\\rm{N}}{{\\rm{a}}^ + }Na+ions.

7 0

3 years ago

Express the following numbers as standard form: a) 1.52 x 10^-2 b) 7.78 x 10^-8

Vikentia [17]

Answer:

0.0152

0.0000000778

Explanation:

You just need to more the decimal as many times as the exponent to the to the left if it is negativa and to the right if it is positive.

4 0

3 years ago

How could you use two beakers distilled water two hot plates two thermometers and salt to test if adding salt affects the boilin

ICE Princess25 [194]

You are given two beakers, distilled water, two hot plates, two thermometers and salt. These materials are enough in order to test the effect of salt in the boiling point water. To do this, you set up two beakers. In one of the beakers, you add pure distilled water and nothing else. For the other beaker, you put a solution of salt and water. You place these beakers on separate hot plates and place inside the beakers the thermometers. You heat these substances until they boil and then you measure the boiling points of the substances. You would observe that the boiling point of the solution would have a higher boiling point than the pure liquid.

3 0

4 years ago

A liquid was analyzed to be 54.5% c, 9.10% h, and 36.4% o. an empty flask, whose mass was 45.32 g, when filled with the vapor of

nataly862011 [7]

number Moles of C = 54.5 g / 12.011 = 4.54 number Moles of H = 9.10 / 1.008 = 9.02 number Moles of O = 36.4 / 16 = 2.28 if we want to divide by the smallest number 4.54 / 2.28 = 2 => C 9.02 / 2.28 = 4 => H 2.28 / 2.28 = 1 => O Empirical formula will be = C2H4O

7 0

4 years ago