Answer:
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
![[OH^{-}] = 10^{-pOH}= 10^{-8.43}= 3.72*10^{-9} [OH^{-}]= 3.72*10^{-9}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%20%3D%2010%5E%7B-pOH%7D%3D%2010%5E%7B-8.43%7D%3D%203.72%2A10%5E%7B-9%7D%0A%0A%20%5BOH%5E%7B-%7D%5D%3D%203.72%2A10%5E%7B-9%7D)
