# Calculate the hydronium ion concentration in 50.0 mL of 0.10M NaH2AsO4.

and we are given K1 = 6.0E-3, K2 = 1.1E-7 and K3 = 3.0E-12

I assumed these K values where given in respect to the Ka of acid H3AsO4.

(H^+) = sqrt[(k2(H2AsO4^-2)+Kw)/1+(H2AsO4^-2/k1)]

I think this can be simplified by calling (HAsO4^-2) = (salt) = C and this reduces to

(H^+) = sqrt[(k2C + kw)/1 + (C/k1)]

Since C/k1 is quite a bit larger than 1, this reduces to

(H^+) = sqrt (k1k2).

This latter one is the one I would use. For references on how this was derived, look in your text on how NaHCO3 is done.

## Oh, calculating hydronium ion concentration, huh? Well, let's get the math party started!

First, let's simplify things by calling (HAsO4^-2) as C. So, we're left with the equation:

(H^+) = sqrt[(k2C + Kw)/1 + (C/k1)]

Now, since C/k1 is quite a bit larger than 1, we can simplify further:

(H^+) = sqrt(k1k2)

Ta-da! This is the equation we'll use to calculate the hydronium ion concentration. So, go ahead and plug in those K values and let the joke-tastic calculation begin!

## To calculate the hydronium ion concentration in the solution, we can use the equation (H^+) = sqrt (k1k2). Here's how you can calculate it step-by-step:

Step 1: Calculate the product of k1 and k2.

k1 = 6.0E-3

k2 = 1.1E-7

k1k2 = (6.0E-3) * (1.1E-7)

Step 2: Take the square root of the product.

(H^+) = sqrt (k1k2)

Hydronium ion concentration = sqrt [(6.0E-3) * (1.1E-7)]

Perform the multiplication and take the square root:

Hydronium ion concentration ≈ 9.13E-5 (rounded to 3 significant figures)

Therefore, the hydronium ion concentration in the given solution is approximately 9.13E-5 M.

## To calculate the hydronium ion concentration in the given solution, we need to use the dissociation constants (K values) provided and the formula you mentioned:

(H^+) = sqrt[(k2C + Kw)/(1 + (C/k1))]

Here, C represents the concentration of the salt NaH2AsO4.

1. Start by calculating the value of C, which represents the concentration of NaH2AsO4 in moles per liter (M).

C = (0.10 M) / (1000 mL/L) * (50.0 mL) = 0.005 mol

2. Substitute the value of C into the formula:

(H^+) = sqrt[(k2 * 0.005 + Kw)/(1 + (0.005/k1))]

3. Substitute the given K values:

(H^+) = sqrt[(1.1E-7 * 0.005 + 1E-14)/(1 + (0.005/6.0E-3))]

4. Simplify the equation further:

(H^+) = sqrt[(5.5E-12 + 1E-14)/(1 + 8.33E-4)]

(H^+) = sqrt[5.51E-12/1.000833]

(H^+) ≈ sqrt[5.5071E-12]

(H^+) ≈ 2.346E-6 M or 2.35E-6 M (rounded to 3 significant figures)

Therefore, the hydronium ion concentration in 50.0 mL of 0.10M NaH2AsO4 is approximately 2.35E-6 M.