Hello!
Data:
Molar Mass of H2CO3 (carbonic acid)
H = 2*1 = 2 amu
C = 1*12 = 12 amu
O = 3*16 = 48 amu
------------------------
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) → [tex] 1*10^{-2}\:M [/tex]
Use: Ka (ionization constant) = [tex] 4.4*10^{-7} [/tex]
[tex] \alpha^2 (degree\:of\:ionization) = ? [/tex]
[tex] Ka = M * \alpha^2 [/tex]
[tex] 4.4*10^{-7} = 1*10^{-2}*\alpha^2 [/tex]
[tex] 1*10^{-2}*\alpha^2 = 4.4*10^{-7} [/tex]
[tex] \alpha^2 = \dfrac{4.4*10^{-7}}{1*10^{-2}} [/tex]
[tex] \alpha^2 = 4.4*10^{-7-(-2)} [/tex]
[tex] \alpha^2 = 4.4*10^{-7+2} [/tex]
[tex] \alpha^2 = 4.4*10^{-5} [/tex]
[tex] \alpha = \sqrt{4.4*10^{-5}} [/tex]
[tex] \boxed{\alpha \approx 2.09*10^{-5}} [/tex]
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:
[tex] [ H_{3} O^+] = M* \alpha [/tex]
[tex] [ H_{3} O^+] = 1*10^{-2}* 2.09*10^{-5} [/tex]
[tex] [ H_{3} O^+] = 2.09*10^{-2-5} [/tex]
[tex] \boxed{[ H_{3} O^+] = 2.09*10^{-7}} [/tex]
And finally, we will use the data found and put in the logarithmic equation of the PH, thus:
Data:
[tex] pH = \:? [/tex]
[tex] [ H_{3} O^+] = 2.09*10^{-7} [/tex]
apply the data to formula
[tex] pH = - log[H_{3} O^+] [/tex]
[tex] pH = - log[2.09*10^{-7}] [/tex]
[tex] pH = 7 - log\:2.09 [/tex]
[tex] pH = 7 - 0.32 [/tex]
[tex] \boxed{pH = 6.68} [/tex]
Note:. The pH <7, then we have an acidic solution (weak acid).
Now, let's find pOH by the following formula:
[tex] pH + pOH = 14 [/tex]
[tex] 6.68 + pOH = 14 [/tex]
[tex] pOH = 14 - 6.68 [/tex]
[tex] \boxed{\boxed{pOH = 7.32}}\end{array}}\qquad\checkmark [/tex]
I Hope this helps, greetings ... DexteR! =)
