A 8.6 X10^-3M Solution of H3PO4 has a Ph=2.30. what is Ka for H3PO4?

Have no clue how to solve this problem

Well, well, well, looks like we've got a tricky chemistry question here. Don't worry, my chemically inclined friend. I'm here to give you a giggle and help you out.

To solve this problem, we need to use the equation for the acidity constant (Ka) of a weak acid. In this case, the weak acid is H3PO4.

Now, let's start by using the equation for pH: pH = -log[H+]. In acidic solutions, [H+] concentration is the same as the concentration of our weak acid (H3PO4). So, pH = -log[H3PO4].

We can rewrite the equation as 10^(-pH) = [H3PO4].

Since the given pH is 2.30, we find that [H3PO4] = 10^(-2.30).

Now, we know that H3PO4 dissociates in water to form H+ and H2PO4-. The dissociation equation looks like this: H3PO4 ⇌ H+ + H2PO4-.

The concentration of H+ is equal to the concentration of our weak acid (H3PO4), so [H+] = [H3PO4].

We also know that [H2PO4-] is equal to the initial concentration of our weak acid (H3PO4), since only a small fraction dissociates.

With that information, we can set up the expression for the acidity constant (Ka) of H3PO4: Ka = [H+][H2PO4-] / [H3PO4].

Substituting our values, Ka = ([H3PO4])([H2PO4-]) / [H3PO4] = [H3PO4].

Therefore, the Ka for H3PO4 is 10^(-2.30).

To solve this problem, you need to understand the relationship between the pH and the concentration of H3PO4, as well as the concept of Ka.

1. Start by understanding the chemical equation for the dissociation of H3PO4 in water:

H3PO4 ⇌ H+ + H2PO4-

The Ka value represents the acid dissociation constant, which is a measure of the strength of an acid. It can be calculated using the following equation:

Ka = [H+][H2PO4-] / [H3PO4]

2. The pH is a measure of the acidity or alkalinity of a solution. In this case, the given pH is 2.30, which is acidic. The pH can be calculated using the equation:

pH = -log[H+]

Rearranging the equation, we find:

[H+] = 10^(-pH)

3. Given that the concentration of H3PO4 is 8.6 X 10^(-3) M, we can substitute these values into the Ka equation:

Ka = [H+][H2PO4-] / [H3PO4]
= (10^(-pH)) * (10^(-pH)) / (8.6 X 10^(-3))

4. Plug in the given pH value of 2.30 into the above equation and perform the calculations:

Ka = (10^(-2.30)) * (10^(-2.30)) / (8.6 X 10^(-3))

Evaluating the numerical values, we find:

Ka ≈ 7.45 X 10^(-8)

Thus, the Ka value for H3PO4 is approximately 7.45 X 10^(-8).

what is the molarity of an aqueous solution containing 40 g of glucose (c6h12o6) in 1.5 L in solution

molarity = moles/L soln.

moles = 40/molar mass.
L soln = 1.5 L
Plug and chug.

H3PO4 ==> H^+ + H2PO4^-

K1 = (H^+)(H2PO4^-)/(H3PO4)

You are given H3PO4 and pH. For all practical purposes, although H3PO4 has k1, k2, and k3, the first ionization of k1 is so large in comparison with the others that you may consider this a monoprotic acid, at least for the purposes of this problem. Use pH to find (H^+), set up and ICE chart, substitute into Ka expression and solve for Ka.