Question

Question #dd23d

Answer 1

The concentration of `"H"^+` is `3.4 × 10^-4"mol/L"`.

Let's set up an ICE table for the calculation.

`K_"a" = (["H"^+]["HCO"_3^-])/(["H"_2"CO"_3]) = (x × x)/(0.25-x) = x^2/(0.25-x) = 4.5 × 10^-7`

`["H"_2"CO"_3]_0/K_a = 0.25/(4.5 × 10^-7) = 5.6 ×10^5`.

Since this is greater than 400, we can say that `x ≪ 0.25`, and the equation becomes

`x^2/0.25 = 4.5 × 10^-7`

`x^2 = 0.25 × 4.5 × 10^-7 =1.12 × 10^-7`

`x = 3.4 × 10^-4`

`["H"^+] = 3.4 × 10^-4 "mol/L"`