Question

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

Answer 1

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

Step-by-step 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

Answer 2

The pH of a 4.3x10^-7 M solution of H2CO3 is approximately 6.4, which indicates that the solution is slightly acidic.

The correct answer is option C.) 6.4.

The pH of a solution is the measure of its acidity or alkalinity, represented by a number on a scale of 0 to 14. To find the pH of a 4.3x10-7 M solution of H2CO3, we can use the formula

pH = -log[H3O+]

Inserting the given hydrogen ion concentration into the formula gives us:

pH = -log[4.3x10-7]

By calculating this value, we find that:

pH ≈ 6.4

Therefore, the correct answer is C.) 6.4, indicating that the solution is slightly acidic.