Question

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

Answer 1

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.

Answer 2

We have that the  pH of the solution  derived to be

$pH=5.0969$

From the Question we are told that

Mass of CH3COONa=1.60g

Volume of CH3COONa v=40ml

0.10 M acetic acid

Ka of CH3COOH is 1.75 × 10^-5.​

Generally the equation for the  pH  is mathematically given as

$pH = pKa + log\frac{ base}{acid}$

Where

$pka = -logKa\\\\pKa = -log(1.75*10^{-5})$

Generally

$Moles\ of\ CH3COONa = \frac{Mass of CH3COONa}{Molar mass of CH3COONa}$

$Moles of CH3COONa=\frac{1.15}{82.03}$

$Moles of CH3COONa=0.0140moles$

And

$base=\frac{ 0.0140moles}{(64.0/1000)}\\\\base=0.21875moles$

Therefore returning to the initial pH equation

$pH = pKa + log\frac{ base}{acid}\\\\pH = -log(1.75*10^{-5}) + log\frac{ 0.21875}{0.10}$

$pH=5.0969$

In conclusion

The  pH of the solution  derived to be

$pH=5.0969$

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