What is the pH of 4.3x10^-7 M solution of H2CO3? A.)7

B.)4.3

C.)6.4

D.)7.6

ASAP PLS

pH = -log[H⁺] = -log(4.75 x 10⁻⁷) = -(- 6.32 ) = 6.32

Explanation:

Given 4.3 x 10⁻⁷M H₂CO₃ (Ka1 = 4.2 x 10⁻⁷ &  Ka2 = 4.8 x 10⁻¹¹)

Note: The Ka2 value for the 2nd ionization step is so small (Ka2 = 4.8 x 10⁻¹¹) It will be assumed all of the hydronium ions (H⁺) come from the 1st ionization step.  

1st Ionization step

                    H₂CO₃        ⇄   H⁺ + HCO₃⁻

C(initial)     4.3 x 10⁻⁷             0         0

ΔC                   -x                  +x        +x

C(final)      4.3 x 10⁻⁷ - x         x          x

Note: the 'x' value in this analysis can not be dropped as the Conc/Ka value is less than 10². In this case he C/Ka ratio* (4.3E-7/4.2E-7 ≈ 1)  is far below 10².

So, one sets up the equilibrium equation to be quadric and the x-value can be determined.

Ka1 = [H⁺][HSO⁻]/[H₂SO₃] = (x)(x)/(4.3 x 10⁻ - x) = x²/(4.3 x 10⁻⁷ - x) = 4.2 x 10⁻⁷

=>   x² = 4.2 x 10⁻⁷(4.3 x 10⁻⁷ - x)

=>   x² + 4.2 x 10⁻⁷x - 1.8 x 10⁻¹³ = 0              

      a = 1, b = 4.2 x 10⁻⁷, c = - 1.8 x 10⁻¹³

x = b² ± SqrRt(b² - 4(1)(-1.8 x 10⁻¹³ / 2(1) =  4.75 x 10⁻⁷

x = [H⁺] = 4.75 x 10⁻⁷

pH = -log[H⁺] = -log(4.75 x 10⁻⁷) = -(- 6.4 ) = 6.4

* The Concentration/Ka-value is the simplification test for quadratic equations used in Equilibrium studies. If the C/Ka > 100 then one can simplify the C(final) by dropping the 'x' if used in this type analysis. However, if the C/Ka value is < 100 then the x-value must be retained and the solution is determined using the quadratic equation formula.

for ax² + bx + c = 0

x = b² ± SqrRt(b² - 4ac) / 2a