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Calculate the pH of a solution prepared by dissolving 1.60 g of sodium acetate, CH3COONa, in 50.0 mL of 0.10 M acetic acid, CH3COOH(aq). Assume the volume change upon dissolving the sodium acetate is negligible. Ka of CH3COOH is 1.75 x 10^-5.
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Sagot :

Drpelezoidn

Answer:

pH = 5.35

Explanation:

Given 1.60 grams sodium acetate (NaOAc(aq))*** added to 50ml of 0.10M acetic acid (HOAc(aq)) solution.

Applying common ion effect keeping in mind that the addition of NaOAc provides the common-ion (OAc⁻).

      HOAc(aq) ⇄          H⁺(aq)            +                 OAc⁻(aq)

 I      0.10m          1.32 x 10⁻³M ≈ ∅M*     (1.6g/82.03g/mol) / 0.050L = 0.39M

 C       -x                         +x                                0.39M + x ≈ 0.39M**

 E    0.10M - x                  x                                    0.39M

       ≈ 0.10M

Ka = [H⁺][OAC⁻]/[HOAC] => [H⁺] = Ka·[HOAc] / [OAc⁻]

[H⁺] = (1.75 X 10⁻⁵)(0.10) / (0.39) = 4.5 x 10⁻⁶M

∴ pH = -log[H⁺] = -log(4.5 x 10⁻⁶) = -(-5.35) = 5.35

_______________________________________________

* [H⁺] before adding NaOAc = SqrRt(Ka · [HOAc]) = SqrRt(1.75 x 10⁻⁵· 0.10) = 1.32 x 10⁻³M. Since this concentration value is so small, the initial [H⁺] is assumed to be zero molar (∅M).

** The added [H⁺] is negligible and dropped in the ICE table. That is, adding ~[H⁺] in the order of 10⁻³M does not change the H⁺ ion concentration sufficiently to affect problem outcome and is therefore dropped in the ICE table.

*** Acetic Acid and Sodium Acetate are frequently written HOAc and NaOAc where the OAc⁻ anion is the acetate ion (CH₃COO⁻) for brevity.

The pH of the solution measures the acid of the liquid throughout the molar concentration of hydrogen ions in a solution.

[tex]\bold{CH_3COONa \ mass = 1.60\ g}\\\\[/tex]

Solution Volume (V) [tex]=50.0\ mL =0.05\ L\\\\[/tex]

[tex]\bold{CH3COONa \ molarity =0.10\ M }[/tex]

[tex]\bold{\text{Calculating the CH3COONa moles} =\frac{mass}{molar\ mass}}[/tex]

                                                       [tex]= \frac{1.60\ g}{82.03\frac{g}{mol}} \\\\=0.0195\ mol\\\\[/tex]

[tex]\bold{M = \frac{ \text{amount of solute moles} } { \text{solution volume in L} }}[/tex]

     [tex]= \frac{0.0195\ mol}{0.05\ L}\\\\ =0.39\ M[/tex]

Using Henderson Hasselbach equation:

[tex]\bold{pH = -\log K_{a} + \log{[salt]}{[acid]}}\\\\[/tex]

     [tex]=\bold{ -\log (1.75\times 10^{-5}) + \log ( \frac{0.39}{0.10}) }\\\\=\bold{ 4.757 + \log (3.9)}\\\\=\bold{ 4.757 + 0.5910}\\\\=\bold{ 5.348}\\\\=\bold{ 5.35}\\[/tex]

So, the final answer is "5.35".

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