# Question dd23d

May 22, 2015

The concentration of ${\text{H}}^{+}$ 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"#