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

a basic solution is an aqueous solution containing more oh-ions than h+ions. in other words, it is an aqueous solution with a ph greater than 7. basic solutions contain ions, conduct electricity, turn red litmus paper blue, and feel slippery to the touch.

explanation:

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be