Chemistry -Acid Weight vs Solution pH

[mindless1]

This might be a simple question for someone good at chemistry.

How many grams of ascorbic acid (C6H8O6) would be needed to lower the pH of 1 liter neutral water to 4.0 ?

Wikipedia lists the acidity of ascorbic acid (pKa) as 4.10 (first), 11.6 (second). Since I have no idea what that means it would help if you showed me how you found the answer.

 

[Paperdoc]

It's been a while since I did this, but here's my recollection.

First, the notation using the lower-case letter p in front of something else means "the negative logarithm (base 10) of ....". So, pH means the negative log of the concentration (in units of molarity, or moles per litre) of Hydrogen ions; a value of pH = 4.0 means that the solution of acid contains 10^(-4) moles pf H+ ions per litre of solution. Similarly, pKa means "the negative log of the value of the acid's Dissociation Constant, Ka".

Now, Ascorbic acid is dibasic - that is, it has two possible hydrogen ions that can dissociate off of the molecule. The first one comes off easily, with pKa = 4.10. That means it will be dissociating readily in acidic environments that already have many H+ ions present. The second one will not come off until the environment has a huge deficiency of H+ ions - that is, it is basic - and its pKa is 11.6. You can ignore the second pKa thing. When you dissolve ascorbic acid in pure distilled water, effectively only the first acidic hydrogen will dissociate off, so work only with that value.

The Dissociation Constant, Ka, is the ratio of the concentrations of product species to source species in the dissociation equilibrium reaction. We will write the ascorbic acid as HA which can dissociate into ions H+ and A-. Note that, when one molecule of HA dissociates, it produces one ion each of H+ and A-. So, let us set up two algebraic variables:

Let x be the initial concentration of HA placed into the solution. Note that the units of concentration are moles per liter (Molarity). x is the amount of dry ascorbic acid powder (in moles) in one litre of pure water before its dissolves and dissociates.

Let y be the number of moles per litre of the original acid that dissociates.

Thus, at equilibrium after dissociation is stable, we have in solution (x-y) moles per liter of undissociated original ascorbic acid, and y moles per litre each of the H+ and A- ions. Then the equation for the Dissociation Constant is:

Ka = y*y/(x-y) = 10^(-4.10)

The second equation to set up comes from your target: the pH of the solution should come out to 4.0. That is, by definition of the term pH, the concentration of the H+ ions in the solution should be 10^(-4.0):

y = 10^(-4.0)

So, solve to get the value for x in moles per litre. Then calculate the Molecular Weight of Ascorbic Acid and convert moles per litre to grams per litre.

 

[mindless1]

OH, is that all?:$ Way over my head. I think I'll get out my pH strips but thanks for the answer anyway! I may just need to read that a few more times.

 

[Paperdoc]

Other than the algebra I set up there, you need to understand these concepts in Chemistry for this problem:

Molarity (a unit of concentration, moles of solute per litre of solution)

The concept of a chemical equilibrium, and the way to calculate the Equilibrium Constant (In this case, the problem concerns a specific type of reaction, a Dissociation Reaction, so the constant is called the Dissociation Constant.) You will note that, in the algebraic expression for Ka, it is calculated as the ratio of the PRODUCT of all the concentrations of the reaction products (on the right-hand side of the reaction equation, and these are H+ ions and A- ions) divided by the PRODUCT of all the concentrations of all the original input species (on the left hand side of the reaction equation, and there is only one here, the undissociated HA molecule).

A Mole (one gram molecular Weight)

A Molecular Weight, and how to get it from the molecular formula

How Logarithms work in mathematics - for this purpose, you only need to work with base 10 logarithms, the most common type. When we write that the "logarithm of a is b", written as:

b = log(a) (where we are assuming that the base of the logs is 10)

what that means in other words is that a is equal to 10 (the base) raised to the power of b. Now, I know that we're not accustomed to evaluating 10 to the power of some decimal fraction or a negative one, but you certainly can do that. The old way was to use logarithm tables - tables of the logarithms of numbers. Today everyone just uses the log function key on a calculator to get the value of log(a). But if you have that value (b), you can find the original value of a by just raising 10 to the power b, and that's also easy on many calculators.

It also helps if you know how to multiply powers of ten terms, like this:

10^g * 10^h = 10^(g+h)

and divide, like this:

10^g / 10^k = 10^(g-k)