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Answer 1 · 2015-05-22T13:37:04

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

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

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

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

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

$x^2/0.25 = 4.5 × 10^-7$

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

$x = 3.4 × 10^-4$

$["H"^+] = 3.4 × 10^-4 "mol/L"$