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.