Chemistry, 14.07.2019 10:00 latoyatuggle23

Determine the poh of 0.01 molar solution of carbonic acid

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Answer from: michellemonchez103

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! =)

Answer from: Quest

Inot sure what it is i sorry : ((

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.