Question

Determine the poh of 0.01 molar solution of carbonic acid

Answer 1

Hello!

Data:

Molar Mass of H2CO3 (carbonic acid)

H = 2*1 = 2 amu

C = 1*12 = 12 amu

O = 3*16 = 48 amu

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Molar Mass of H2CO3 = 2 + 12 + 48 = 62 g/mol

Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:

M (molarity) = 0.01 M (Mol/L) → $1*10^{-2}\:M$

Use: Ka (ionization constant) = $4.4*10^{-7}$

$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$

$4.4*10^{-7} = 1*10^{-2}*\alpha^2$

$1*10^{-2}*\alpha^2 = 4.4*10^{-7}$

$\alpha^2 = \dfrac{4.4*10^{-7}}{1*10^{-2}}$

$\alpha^2 = 4.4*10^{-7-(-2)}$

$\alpha^2 = 4.4*10^{-7+2}$

$\alpha^2 = 4.4*10^{-5}$

$\alpha = \sqrt{4.4*10^{-5}}$

$\boxed{\alpha \approx 2.09*10^{-5}}$

Now, we will calculate the amount of Hydronium [H3O+] in carbonic acid (H2CO3), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$

$[ H_{3} O^+] = 1*10^{-2}* 2.09*10^{-5}$

$[ H_{3} O^+] = 2.09*10^{-2-5}$

$\boxed{[ H_{3} O^+] = 2.09*10^{-7}}$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:

$pH = \:?$

$[ H_{3} O^+] = 2.09*10^{-7}$

apply the data to formula

$pH = - log[H_{3} O^+]$

$pH = - log[2.09*10^{-7}]$

$pH = 7 - log\:2.09$

$pH = 7 - 0.32$

$\boxed{pH = 6.68}$

Note:. The pH <7, then we have an acidic solution (weak acid).

Now, let's find pOH by the following formula:

$pH + pOH = 14$

$6.68 + pOH = 14$

$pOH = 14 - 6.68$

$\boxed{\boxed{pOH = 7.32}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)