calculate oh- and ph in a solution in which dihydrogen phosphate ion is 0.335 M and hydrogen phosphate is 0.335 M.

This is a homework question, where I have the solution, which is 0.335/0.335, which I get, but then I am supposed to multiply this (1)by a number 6.2x10-8? I have absolutely no clue where this number is coming from. What is this number? How would I find it?

http://en.wikipedia.org/wiki/Phosphate You are looking at Ka2

You need to write the dissociation equation, and the expression for solubility constant.

I am not sure I understand what you are saying, but the number that number seems to be the Ka value for phosphoric acid.

pH=pka+log[A^-/HA]

let

pka=-log[Ka]
Ka=6.2 x 10^-8
pka=-log[Ka]=7.21
A^-=0.335
HA=0.335

To help you out a little bit:
The concentrations are equal, so the pka=pH

that is, the pH=7.21

pH+pOH=14

So, you can use that to solve for the pOH

pOH=-log[OH^-]

So, if you need the [OH^-] concentrations, use the above equation.

10^-(pOH)=[OH^-]

The number you mentioned, 6.2x10^-8, is the equilibrium constant (Ka) for the dissociation reaction of the dihydrogen phosphate ion (H2PO4^-). Let's go through the calculations step by step:

1. Write the dissociation equation for dihydrogen phosphate ion:
H2PO4^- ⇌ HPO4^2- + H+

2. Construct the equilibrium expression using the concentrations of the species:
Ka = [HPO4^2-][H+] / [H2PO4^-]

3. Substitute the given concentrations into the equation. Since the concentrations of dihydrogen phosphate ion and hydrogen phosphate are the same (0.335 M), we can assume that the initial concentration of H2PO4^- is equal to its change in concentration ([H2PO4^-] = -Δ[H2PO4^-]), so [H2PO4^-] in the equilibrium expression can be represented as (0.335 - x), where x is the concentration of H+.

Ka = [HPO4^2-][H+] / (0.335 - x)

4. Since the dissociation of H2PO4^- produces one H+ ion and one HPO4^2- ion, at equilibrium, the concentration of H+ is equal to the concentration of HPO4^2-. Therefore, we can replace [HPO4^2-] in the equation with (0.335 - x):

Ka = (0.335 - x)(0.335 - x) / (0.335 - x)

5. Simplify the equation:
Ka = (0.335 - x)(0.335 - x) / 0.335

6. Rearrange the equation to solve for x:
(0.335 - x)(0.335 - x) = Ka * 0.335

7. Solve for x. This requires the use of the quadratic equation, and the value of x will be the concentration of H+:
x = (0.335 - sqrt(Ka * 0.335))

8. Once you calculate the value of x, substitute it into the equation to find the pH:
pH = -log[H+]

The value of 6.2x10^-8 is the specific Ka value given in the question. If this value was not provided, you would need to look it up in a reference source or calculate it from thermodynamic data.

To calculate the OH- and pH in a solution containing dihydrogen phosphate (H2PO4-) and hydrogen phosphate (HPO42-), you need to consider the equilibrium reaction between these two species:

H2PO4- ⇌ HPO42- + H+

The reaction can be described by an equilibrium constant, commonly known as the acid dissociation constant (Ka):
Ka = [HPO42-][H+]/[H2PO4-]

To simplify the calculations, we will assume that the concentration of H+ ions is equal to the concentration of OH- ions, which will be the case for neutral solutions.

Now, let's solve the problem step-by-step:

1. Find the value of [HPO42-]/[H2PO4-]:
Given that both [H2PO4-] and [HPO42-] are 0.335 M, the ratio [HPO42-]/[H2PO4-] is 0.335/0.335 = 1.

2. Calculate the OH- ion concentration ([OH-]):
Since we assumed a neutral solution, the concentration of OH- ions is equal to the concentration of H+ ions. Therefore, [OH-] is equal to [H+].

3. Determine the value of Ka:
The problem states that you need to multiply the ratio [HPO42-]/[H2PO4-] by a specific constant, 6.2x10-8. This constant is the Ka for the reaction. It indicates the strength of the acid.

4. Solve for [OH-]:
To find [OH-], first, you need to find [H+] using the equilibrium expression:
Ka = [HPO42-][H+]/[H2PO4-]

Since we assumed that [H+] = [OH-], the equation can be rewritten as:
Ka = ([HPO42-][OH-])/[H2PO4-]

Rearranging this equation to solve for [OH-], we get:
[OH-] = (Ka x [H2PO4-])/[HPO42-]

Inserting the values we have:
[OH-] = (6.2x10-8 x 0.335)/1 = 2.07x10-8 M

5. Calculate the pH:
To determine the pH, we need to take the negative logarithm (base 10) of the [H+] concentration:
pH = -log10([H+]) = -log10([OH-])

Given that [OH-] = 2.07x10-8 M, we can find the pH as follows:
pH = -log10(2.07x10-8) = 7.68

Therefore, in the given solution, the concentration of OH- is 2.07x10-8 M, and the pH is 7.68.

Note: The constant 6.2x10-8 represents the Ka value for the acid dissociation reaction specific to dihydrogen phosphate and hydrogen phosphate ions. This constant is specific to this particular acid-base system and is typically provided in context or given in reference tables.