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How many grams of $\ce{CH3COONa}$ we need to dissolve in $\pu{2.50 L}$ of water to get a solution with $\mathrm{pH} = 9.00$? Given $K_\mathrm{a}(\ce{CH3COOH}) = 1.75\cdot 10^{-5}$

Like solving a puzzle, I found from the asked $\mathrm{pH}$ the concentration of hydrogen cation and hydroxide anion, and wrote to myself that I need to find the final concentration of $\ce{CH3COONa}$ to seek from it the amount and then the weight.

But my problem was in all the middle steps. I understand the the concentration of hydroxide anions that I found is after the equilibrium, but I miss the point of how to use them to find the concentration on the acetic acid by given the $K_\mathrm{a}$. I'm guessing that I need to get the $K_\mathrm{b}$ from the $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$ equation. But why exactly and how then, I need to find the concentration of $\ce{CH3COONa}$?

Maybe I lack the knowledge and thus the imagination for it.

Support highly appreciated!

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It's not really a matter of remembering a certain formula, this is not a conceptual thinking and should be avoided. When approaching these types of questions, always start with a balanced chemical equation. In this case we deal with a salt formed by a weak acid and a strong base, which hydrolyses according to the following net ionic equation:

$$\ce{CH3COO- + H2O <=>[$K_\mathrm{h}$] CH3COOH + OH-}$$

$$ \begin{align} K_\mathrm{h} &= \left.\frac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}\qquad\right|\cdot\frac{[\ce{H+}]}{[\ce{H+}]}\\ &= \frac{[\ce{CH3COOH}][\ce{OH-}][\ce{H+}]}{[\ce{CH3COO-}][\ce{H+}]}\\ &= \frac{K_\mathrm{w}}{K_\mathrm{a}}\label{eq:1}\tag{1} \end{align} $$

At the same time disregarding autoprotolysis of water, and taking into account that $[\ce{CH3COOH}]\approx [\ce{OH-}]$ and $[\ce{CH3COO-}]\approx c$ ($c$ is the concentration of sodium acetate), hydrolysis constant is

$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{c}\label{eq:2}\tag{2}$$

Knowing that

$$[\ce{OH-}] = \frac{K_\mathrm{w}}{[\ce{H+}]} = \frac{K_\mathrm{w}}{10^{-\mathrm{pH}}}\label{eq:3}\tag{3}$$

one can equate \eqref{eq:1} and \eqref{eq:2}, substitute $[\ce{OH-}]$ with \eqref{eq:3}, and express concentration as

$$ \require{cancel} c = \frac{K_\mathrm{a}[\ce{OH-}]^2}{K_\mathrm{w}} = \frac{K_\mathrm{a}K_\mathrm{w}^{\cancel{2}1}}{10^{-2\mathrm{pH}}\cancel{K_\mathrm{w}}} = \frac{K_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}}\tag{4} $$

Finally, mass $m$ of sodium acetate can be found from its molar mass $M$ and volume of the solution $V$:

$$m = cVM = \frac{VMK_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}} = \frac{\pu{2.50 L}\cdot\pu{82.03 g mol-1}\cdot 1.75\cdot 10^{-5} \cdot 10^{-14}}{\pu{10^{-2\cdot 9} mol-1 L}} = \pu{35.89 g}\tag{5}$$

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  • $\begingroup$ First of all thanks for your explanation, im learning a lot her in the community. I understood setps 1,2 and 4 but step 3 i`m a little bit confused. How did the Kw become $\ce{10^{-2pH}}$ in the denominator? $\endgroup$ – Mabadai Jan 5 '18 at 21:44
  • $\begingroup$ @Mabadai And thank you for your feedback. I tried to explain the operations behind former equation (3), hopefully now it's somewhat clearer. $\endgroup$ – andselisk Jan 5 '18 at 22:22
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pH = 0.5(pKa + logC) pKa is log of Ka

C is the concentration

pH is Given 9.

Hence,

9= 0.5[(log1.75) - 5 + logC]

18= 0.2430 - 5 + logC

22.75 = LogC

This equation will give you Molarity.

Use molecular mass of Ch3COONa to convert it to grams

I may be wrong but I think I've used the right formula

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  • $\begingroup$ Your answer is fine. Because i need more than "just using a formula by a small prediction". I want to understand things, and learn them from the real point of my life. Solving those questions is good for exam and so on, but i dont want to forget them after i pass the test. Thats why i`m seeking for more explanation and logic than some tricky math and equations. $\endgroup$ – Mabadai Jan 5 '18 at 21:48

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