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

------------------------

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 = (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! =)